In Chapter 3, we discovered functions as a way to *abstract* expressions; to
rephrase a particular computation with some parts missing. We used functions to
transform a single value. But what if we want to apply a function to *more than
one* value at once? What about sequences?

For example, we know that `(inc 2)`

increments the number 2. What if we wanted to increment *every number* in the vector `[1 2 3]`

, producing `[2 3 4]`

?

```
user=> (inc [1 2 3])
ClassCastException clojure.lang.PersistentVector cannot be cast to java.lang.Number clojure.lang.Numbers.inc (Numbers.java:110)
```

Clearly `inc`

can only work on numbers, not on vectors. We need a different
kind of tool.

## A direct approach

Let’s think about the problem in concrete terms. We want to increment each of three elements: the first, second, and third. We know how to get an element from a sequence by using nth, so let’s start with the first number, at index 0:

```
user=> (def numbers [1 2 3])
#'user/numbers
user=> (nth numbers 0)
1
user=> (inc (nth numbers 0))
2
```

So there’s the first element incremented. Now we can do the second:

```
user=> (inc (nth numbers 1))
3
user=> (inc (nth numbers 2))
4
```

And it should be straightforward to combine these into a vector…

```
user=> [(inc (nth numbers 0)) (inc (nth numbers 1)) (inc (nth numbers 2))]
[2 3 4]
```

Success! We’ve incremented each of the numbers in the list! How about a list with only two elements?

```
user=> (def numbers [1 2])
#'user/numbers
user=> [(inc (nth numbers 0)) (inc (nth numbers 1)) (inc (nth numbers 2))]
IndexOutOfBoundsException clojure.lang.PersistentVector.arrayFor (PersistentVector.java:107)
```

Shoot. We tried to get the element at index 2, but *couldn’t*, because
`numbers`

only has indices 0 and 1. Clojure calls that “index out of bounds”.

We could just leave off the third expression in the vector; taking only
elements 0 and 1. But the problem actually gets much worse, because we’d need
to make this change *every* time we wanted to use a different sized vector. And
what of a vector with 1000 elements? We’d need 1000 `(inc (nth numbers ...))`

expressions! Down this path lies madness.

Let’s back up a bit, and try a slightly smaller problem.

## Recursion

What if we just incremented the *first* number in the vector? How would that
work? We know that `first`

finds the first element in a sequence, and `rest`

finds all the remaining ones.

```
user=> (first [1 2 3])
1
user=> (rest [1 2 3])
(2 3)
```

So there’s the *pieces* we’d need. To glue them back together, we can use a
function called `cons`

, which says “make a list beginning with the first
argument, followed by all the elements in the second argument”.

```
user=> (cons 1 [2])
(1 2)
user=> (cons 1 [2 3])
(1 2 3)
user=> (cons 1 [2 3 4])
(1 2 3 4)
```

OK so we can split up a sequence, increment the first part, and join them back together. Not so hard, right?

```
(defn inc-first [nums]
(cons (inc (first nums))
(rest nums)))
user=> (inc-first [1 2 3 4])
(2 2 3 4)
```

Hey, there we go! First element changed. Will it work with any length list?

```
user=> (inc-first [5])
(6)
user=> (inc-first [])
NullPointerException clojure.lang.Numbers.ops (Numbers.java:942)
```

Shoot. We can’t increment the first element of this empty vector, because it
doesn’t *have* a first element.

```
user=> (first [])
nil
user=> (inc nil)
NullPointerException clojure.lang.Numbers.ops (Numbers.java:942)
```

So there are really *two* cases for this function. If there is a first element
in `nums`

, we’ll increment it as normal. If there’s *no* such element, we’ll
return an empty list. To express this kind of conditional behavior, we’ll use a
Clojure special form called `if`

:

```
user=> (doc if)
-------------------------
if
(if test then else?)
Special Form
Evaluates test. If not the singular values nil or false,
evaluates and yields then, otherwise, evaluates and yields else. If
else is not supplied it defaults to nil.
Please see http://clojure.org/special_forms#if
```

To confirm our intuition:

```
user=> (if true :a :b)
:a
user=> (if false :a :b)
:b
```

Seems straightforward enough.

```
(defn inc-first [nums]
(if (first nums)
; If there's a first number, build a new list with cons
(cons (inc (first nums))
(rest nums))
; If there's no first number, just return an empty list
(list)))
user=> (inc-first [])
()
user=> (inc-first [1 2 3])
(2 2 3)
```

Success! Now we can handle *both* cases: empty sequences, and sequences with
things in them. Now how about incrementing that *second* number? Let’s stare at
that code for a bit.

```
(rest nums)
```

Hang on. That list–`(rest nums)`

–that’s a list of numbers too. What if we…
used our inc-first function on *that* list, to increment *its* first number?
Then we’d have incremented both the first *and* the second element.

```
(defn inc-more [nums]
(if (first nums)
(cons (inc (first nums))
(inc-more (rest nums)))
(list)))
user=> (inc-more [1 2 3 4])
(2 3 4 5)
```

Odd. That didn’t just increment the first two numbers. It incremented *all* the
numbers. We fell into the *complete* solution entirely by accident. What
happened here?

Well first we… yes, we got the number one, and incremented it. Then we stuck
that onto `(inc-first [2 3 4])`

, which got two, and incremented it. Then we stuck
that two onto `(inc-first [3 4])`

, which got three, and then we did the same for
four. Only *that* time around, at the very end of the list, `(rest [4])`

would
have been *empty*. So when we went to get the first number of the empty list,
we took the *second* branch of the `if`

, and returned the empty list.

Having reached the *bottom* of the function calls, so to speak, we zip back up
the chain. We can imagine this function turning into a long string of `cons`

calls, like so:

```
(cons 2 (cons 3 (cons 4 (cons 5 '()))))
(cons 2 (cons 3 (cons 4 '(5))))
(cons 2 (cons 3 '(4 5)))
(cons 2 '(3 4 5))
'(2 3 4 5)
```

This technique is called *recursion*, and it is a fundamental principle in
working with collections, sequences, trees, or graphs… any problem which has
small parts linked together. There are two key elements in a recursive program:

- Some part of the problem which has a known solution
- A relationship which connects one part of the problem to the next

Incrementing the elements of an empty list returns the empty list. This is our
*base case*: the ground to build on. Our *inductive* case, also called the
*recurrence relation*, is how we broke the problem up into incrementing the
*first* number in the sequence, and incrementing all the numbers in the
*rest* of the sequence. The `if`

expression bound these two cases together into
a single function; a function *defined in terms of itself*.

Once the initial step has been taken, *every* step can be taken.

```
user=> (inc-more [1 2 3 4 5 6 7 8 9 10 11 12])
(2 3 4 5 6 7 8 9 10 11 12 13)
```

This is the beauty of a recursive function; folding an unbounded stream of computation over and over, onto itself, until only a single step remains.

## Generalizing from inc

We set out to increment every number in a vector, but nothing in our solution
actually depended on `inc`

. It just as well could have been `dec`

, or `str`

, or
`keyword`

. Let’s *parameterize* our `inc-more`

function to use *any*
transformation of its elements:

```
(defn transform-all [f xs]
(if (first xs)
(cons (f (first xs))
(transform-all f (rest xs)))
(list)))
```

Because we could be talking about *any* kind of sequence, not just numbers,
we’ve named the sequence `xs`

, and its first element `x`

. We also take a
function `f`

as an argument, and that function will be applied to each `x`

in
turn. So not only can we increment numbers…

```
user=> (transform-all inc [1 2 3 4])
(2 3 4 5)
```

…but we can turn strings to keywords…

```
user=> (transform-all keyword ["bell" "hooks"])
(:bell :hooks)
```

…or wrap every element in a list:

```
user=> (transform-all list [:codex :book :manuscript])
((:codex) (:book) (:manuscript))
```

In short, this function expresses a sequence in which each element is some
function applied to the corresponding element in the underlying sequence.
This idea is so important that it has its own name, in mathematics, Clojure, and other languages. We call it `map`

.

```
user=> (map inc [1 2 3 4])
(2 3 4 5)
```

You might remember maps as a datatype in Clojure, too–they’re dictionaries that relate keys to values.

```
{:year 1969
:event "moon landing"}
```

The *function* `map`

relates one sequence to another. The `type`

map relates
keys to values. There is a deep symmetry between the two: maps are usually
sparse, and the relationships between keys and values may be arbitrarily
complex. The map function, on the other hand, usually expresses the *same* type
of relationship, applied to a series of elements in *fixed order*.

## Building sequences

Recursion can do more than just `map`

. We can use it to expand a single value
into a sequence of values, each related by some function. For
instance:

```
(defn expand [f x count]
(if (pos? count)
(cons x (expand f (f x) (dec count)))))
```

Our base case is `x`

itself, followed by the sequence beginning with `(f x)`

.
That sequence in turn expands to `(f (f x))`

, and then `(f (f (f x)))`

, and so on.
Each time we call `expand`

, we count down by one using `dec`

. Once the count is zero,
the `if`

returns `nil`

, and evaluation stops. If we start with the number 0 and
use inc as our function:

```
user=> user=> (expand inc 0 10)
(0 1 2 3 4 5 6 7 8 9)
```

Clojure has a more general form of this function, called `iterate`

.

```
user=> (take 10 (iterate inc 0))
(0 1 2 3 4 5 6 7 8 9)
```

Since this sequence is *infinitely* long, we’re using `take`

to select only the
first 10 elements. We can construct more complex sequences by using more
complex functions:

```
user=> (take 10 (iterate (fn [x] (if (odd? x) (+ 1 x) (/ x 2))) 10))
(10 5 6 3 4 2 1 2 1 2)
```

Or build up strings:

```
user=> (take 5 (iterate (fn [x] (str x "o")) "y"))
("y" "yo" "yoo" "yooo" "yoooo")
```

`iterate`

is extremely handy for working with infinite sequences, and has some
partners in crime. `repeat`

, for instance, constructs a sequence where every
element is the same.

```
user=> (take 10 (repeat :hi))
(:hi :hi :hi :hi :hi :hi :hi :hi :hi :hi)
user=> (repeat 3 :echo)
(:echo :echo :echo)
```

And its close relative `repeatedly`

simply calls a function `(f)`

to generate an
infinite sequence of values, over and over again, without any relationship
between elements. For an infinite sequence of random numbers:

```
user=> (rand)
0.9002678382322784
user=> (rand)
0.12375594203332863
user=> (take 3 (repeatedly rand))
(0.44442397843046755 0.33668691162169784 0.18244875487846746)
```

Notice that calling `(rand)`

returns a different number each time. We say that
`rand`

is an *impure* function, because it cannot simply be replaced by the
same value every time. It does something different each time it’s called.

There’s another very handy sequence function specifically for numbers: `range`

,
which generates a sequence of numbers between two points. `(range n)`

gives n
successive integers starting at 0. `(range n m)`

returns integers from n to
m-1. `(range n m step)`

returns integers from n to m, but separated by `step`

.

```
user=> (range 5)
(0 1 2 3 4)
user=> (range 2 10)
(2 3 4 5 6 7 8 9)
user=> (range 0 100 5)
(0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95)
```

To extend a sequence by repeating it forever, use `cycle`

:

```
user=> (take 10 (cycle [1 2 3]))
(1 2 3 1 2 3 1 2 3 1)
```

## Transforming sequences

Given a sequence, we often want to find a *related* sequence. `map`

, for
instance, applies a function to each element–but has a few more tricks up its
sleeve.

```
user=> (map (fn [n vehicle] (str "I've got " n " " vehicle "s"))
[0 200 9]
["car" "train" "kiteboard"])
("I've got 0 cars" "I've got 200 trains" "I've got 9 kiteboards")
```

If given multiple sequences, `map`

calls its function with one element from each sequence in turn. So the first value will be `(f 0 "car")`

, the second ```
(f 200
"train")
```

, and so on. Like a zipper, map folds together corresponding elements
from multiple collections. To sum three vectors, column-wise:

```
user=> (map + [1 2 3]
[4 5 6]
[7 8 9])
(12 15 18)
```

If one sequence is bigger than another, map stops at the end of the smaller one. We can exploit this to combine finite and infinite sequences. For example, to number the elements in a vector:

```
user=> (map (fn [index element] (str index ". " element))
(iterate inc 0)
["erlang" "ruby" "haskell"])
("0. erlang" "1. ruby" "2. haskell")
```

Transforming elements together with their indices is so common that Clojure has
a special function for it: `map-indexed`

:

```
user=> (map-indexed (fn [index element] (str index ". " element))
["erlang" "ruby" "haskell"])
("0. erlang" "1. ruby" "2. haskell")
```

You can also tack one sequence onto the end of another, like so:

```
user=> (concat [1 2 3] [:a :b :c] [4 5 6])
(1 2 3 :a :b :c 4 5 6)
```

Another way to combine two sequences is to riffle them together, using
`interleave`

.

```
user=> (interleave [:a :b :c] [1 2 3])
(:a 1 :b 2 :c 3)
```

And if you want to insert a specific element between each successive pair in a
sequence, try `interpose`

:

```
user=> (interpose :and [1 2 3 4])
(1 :and 2 :and 3 :and 4)
```

To reverse a sequence, use `reverse`

.

```
user=> (reverse [1 2 3])
(3 2 1)
user=> (reverse "woolf")
(\f \l \o \o \w)
```

Strings are sequences too! Each element of a string is a *character*, written `\f`

. You can rejoin those characters into a string with `apply str`

:

```
user=> (apply str (reverse "woolf"))
"floow"
```

…and break strings up into sequences of chars with `seq`

.

```
user=> (seq "sato")
(\s \a \t \o)
```

To randomize the order of a sequence, use `shuffle`

.

```
user=> (shuffle [1 2 3 4])
[3 1 2 4]
user=> (apply str (shuffle (seq "abracadabra")))
"acaadabrrab"
```

## Subsequences

We’ve already seen `take`

, which selects the first n elements. There’s also
`drop`

, which removes the first n elements.

```
user=> (range 10)
(0 1 2 3 4 5 6 7 8 9)
user=> (take 3 (range 10))
(0 1 2)
user=> (drop 3 (range 10))
(3 4 5 6 7 8 9)
```

And for slicing apart the other end of the sequence, we have `take-last`

and `drop-last`

:

```
user=> (take-last 3 (range 10))
(7 8 9)
user=> (drop-last 3 (range 10))
(0 1 2 3 4 5 6)
```

`take-while`

and `drop-while`

work just like `take`

and `drop`

, but use a function to decide when to cut.

```
user=> (take-while pos? [3 2 1 0 -1 -2 10])
(3 2 1)
```

In general, one can cut a sequence in twain by using `split-at`

, and giving it
a particular index. There’s also `split-with`

, which uses a function to decide
when to cut.

```
(split-at 4 (range 10))
[(0 1 2 3) (4 5 6 7 8 9)]
user=> (split-with number? [1 2 3 :mark 4 5 6 :mark 7])
[(1 2 3) (:mark 4 5 6 :mark 7)]
```

Notice that because indexes start at zero, sequence functions tend to have
predictable numbers of elements. `(split-at 4)`

yields *four* elements in the
first collection, and ensures the second collection *begins at index four*.
`(range 10)`

has ten elements, corresponding to the first ten indices in a
sequence. `(range 3 5)`

has two (since 5 - 3 is two) elements. These choices simplify the
definition of recursive functions as well.

We can select particular elements from a sequence by applying a function. To
find all positive numbers in a list, use `filter`

:

```
user=> (filter pos? [1 5 -4 -7 3 0])
(1 5 3)
```

`filter`

looks at each element in turn, and includes it in the resulting
sequence *only* if `(f element)`

returns a truthy value. Its complement is
`remove`

, which only includes those elements where `(f element)`

is `false`

or `nil`

.

```
user=> (remove string? [1 "turing" :apple])
(1 :apple)
```

Finally, one can group a sequence into chunks using `partition`

,
`partition-all`

, or `partition-by`

. For instance, one might group alternating
values into pairs:

```
user=> (partition 2 [:cats 5 :bats 27 :crocodiles 0])
((:cats 5) (:bats 27) (:crocodiles 0))
```

Or separate a series of numbers into negative and positive runs:

```
(user=> (partition-by neg? [1 2 3 2 1 -1 -2 -3 -2 -1 1 2])
((1 2 3 2 1) (-1 -2 -3 -2 -1) (1 2))
```

## Collapsing sequences

After transforming a sequence, we often want to collapse it in some way; to derive some smaller value. For instance, we might want the number of times each element appears in a sequence:

```
user=> (frequencies [:meow :mrrrow :meow :meow])
{:meow 3, :mrrrow 1}
```

Or to group elements by some function:

```
user=> (pprint (group-by :first [{:first "Li" :last "Zhou"}
{:first "Sarah" :last "Lee"}
{:first "Sarah" :last "Dunn"}
{:first "Li" :last "O'Toole"}]))
{"Li" [{:last "Zhou", :first "Li"} {:last "O'Toole", :first "Li"}],
"Sarah" [{:last "Lee", :first "Sarah"} {:last "Dunn", :first "Sarah"}]}
```

Here we’ve taken a sequence of people with first and last names, and used the
`:first`

keyword (which can act as a function!) to look up those first names.
`group-by`

used that function to produce a *map* of first names to lists of
people–kind of like an index.

In general, we want to *combine* elements together in some way, using a
function. Where `map`

treated each element independently, reducing a sequence
requires that we bring some information along. The most general way to collapse a sequence is `reduce`

.

```
user=> (doc reduce)
-------------------------
clojure.core/reduce
([f coll] [f val coll])
f should be a function of 2 arguments. If val is not supplied,
returns the result of applying f to the first 2 items in coll, then
applying f to that result and the 3rd item, etc. If coll contains no
items, f must accept no arguments as well, and reduce returns the
result of calling f with no arguments. If coll has only 1 item, it
is returned and f is not called. If val is supplied, returns the
result of applying f to val and the first item in coll, then
applying f to that result and the 2nd item, etc. If coll contains no
items, returns val and f is not called.
```

That’s a little complicated, so we’ll start small. We need a function, `f`

,
which combines successive elements of the sequence. `(f state element)`

will
return the state for the *next* invocation of `f`

. As `f`

moves along the
sequence, it carries some changing state with it. The final state is the return
value of `reduce`

.

```
user=> (reduce + [1 2 3 4])
10
```

`reduce`

begins by calling `(+ 1 2)`

, which yields the state `3`

. Then it calls
`(+ 3 3)`

, which yields `6`

. Then `(+ 6 4)`

, which returns `10`

. We’ve taken a
function over *two* elements, and used it to combine *all* the elements. Mathematically, we could write:

```
1 + 2 + 3 + 4
3 + 3 + 4
6 + 4
10
```

So another way to look at `reduce`

is like sticking a function *between* each
pair of elements. To see the reducing process in action, we can use
`reductions`

, which returns a sequence of all the intermediate states.

```
user=> (reductions + [1 2 3 4])
(1 3 6 10)
```

Oftentimes we include a *default* state to start with. For instance, we could
start with an empty set, and add each element to it as we go along:

```
user=> (reduce conj #{} [:a :b :b :b :a :a])
#{:a :b}
```

Reducing elements into a collection has its own name: `into`

. We can conj ```
[key
value]
```

vectors into a map, for instance, or build up a list:

```
user=> (into {} [[:a 2] [:b 3]])
{:a 2, :b 3}
user=> (into (list) [1 2 3 4])
(4 3 2 1)
```

Because elements added to a list appear at the *beginning*, not the end, this
expression reverses the sequence. Vectors `conj`

onto the end, so to emit the
elements in order, using `reduce`

, we might try:

```
user=> (reduce conj [] [1 2 3 4 5])
(reduce conj [] [1 2 3 4 5])
[1 2 3 4 5]
```

Which brings up an interesting thought: this looks an awful lot like `map`

. All that’s missing is some kind of transformation applied to each element.

```
(defn my-map [f coll]
(reduce (fn [output element]
(conj output (f element)))
[]
coll))
user=> (my-map inc [1 2 3 4])
[2 3 4 5]
```

Huh. `map`

is just a special kind of `reduce`

. What about, say, `take-while`

?

```
(defn my-take-while [f coll]
(reduce (fn [out elem]
(if (f elem)
(conj out elem)
(reduced out)))
[]
coll))
```

We’re using a special function here, `reduced`

, to indicate that we’ve
completed our reduction *early* and can skip the rest of the sequence.

```
user=> (my-take-while pos? [2 1 0 -1 0 1 2])
[2 1]
```

`reduce`

really is the uberfunction over sequences. Almost any
operation on a sequence can be expressed in terms of a reduce–though for
various reasons, many of the Clojure sequence functions are not written this
way. For instance, `take-while`

is *actually* defined like so:

```
user=> (source take-while)
(defn take-while
"Returns a lazy sequence of successive items from coll while
(pred item) returns true. pred must be free of side-effects."
{:added "1.0"
:static true}
[pred coll]
(lazy-seq
(when-let [s (seq coll)]
(when (pred (first s))
(cons (first s) (take-while pred (rest s)))))))
```

There’s a few new pieces here, but the structure is *essentially* the same as
our initial attempt at writing `map`

. When the predicate matches the first
element, cons the first element onto `take-while`

, applied to the rest of the
sequence. That `lazy-seq`

construct allows Clojure to compute this sequence *as
required*, instead of right away. It defers execution to a later time.

Most of Clojure’s sequence functions are lazy. They don’t do anything until needed. For instance, we can increment every number from zero to infinity:

```
user=> (def infseq (map inc (iterate inc 0)))
#'user/infseq
user=> (realized? infseq)
false
```

That function returned *immediately*. Because it hasn’t done any work yet, we
say the sequence is *unrealized*. It doesn’t increment any numbers at all until
we ask for them:

```
user=> (take 10 infseq)
(1 2 3 4 5 6 7 8 9 10)
user=> (realized? infseq)
true
```

Lazy sequences also *remember* their contents, once evaluated, for faster
access.

## Putting it all together

We’ve seen how recursion generalizes a function over *one* thing into a
function over *many* things, and discovered a rich landscape of recursive
functions over sequences. Now let’s use our knowledge of sequences to solve a
more complex problem: find the sum of the products of consecutive pairs of the
first 1000 odd integers.

First, we’ll need the integers. We can start with 0, and work our way up to infinity. To save time printing an infinite number of integers, we’ll start with just the first 10.

```
user=> (take 10 (iterate inc 0))
(0 1 2 3 4 5 6 7 8 9)
```

Now we need to find only the ones which are odd. Remember, `filter`

pares down
a sequence to only those elements which pass a test.

```
user=> (take 10 (filter odd? (iterate inc 0)))
(1 3 5 7 9 11 13 15 17 19)
```

For consecutive pairs, we want to take `[1 3 5 7 ...]`

and find a sequence like `([1 3] [3 5] [5 7] ...)`

. That sounds like a job for `partition`

:

```
user=> (take 3 (partition 2 (filter odd? (iterate inc 0))))
((1 3) (5 7) (9 11))
```

Not quite right–this gave us non-overlapping pairs, but we wanted overlapping
ones too. A quick check of `(doc partition)`

reveals the `step`

parameter:

```
user=> (take 3 (partition 2 1 (filter odd? (iterate inc 0))))
((1 3) (3 5) (5 7))
```

Now we need to find the product for each pair. Given a pair, multiply the two
pieces together… yes, that sounds like `map`

:

```
user=> (take 3 (map (fn [pair] (* (first pair) (second pair)))
(partition 2 1 (filter odd? (iterate inc 0)))))
(3 15 35)
```

Getting a bit unwieldy, isn’t it? Only one final step: sum all those products.
We’ll adjust the `take`

to include the first 1000, not the first 3, elements.

```
user=> (reduce +
(take 1000
(map (fn [pair] (* (first pair) (second pair)))
(partition 2 1
(filter odd?
(iterate inc 0)))))
1335333000
```

The sum of the first thousand products of consecutive pairs of the odd integers
starting at 0. See how each part leads to the next? This expression looks a lot
like the way we phrased the problem in English–but both English and Lisp
expressions are sort of backwards, in a way. The part that *happens first*
appears *deepest*, *last*, in the expression. In a chain of reasoning like
this, it’d be nicer to write it in order.

```
user=> (->> 0
(iterate inc)
(filter odd?)
(partition 2 1)
(map (fn [pair]
(* (first pair) (second pair))))
(take 1000)
(reduce +))
1335333000
```

Much easier to read: now everything flows in order, from top to bottom, and we’ve flattened out the deeply nested expressions into a single level. This is how object-oriented languages structure their expressions: as a chain of function invocations, each acting on the previous value.

But how is this possible? Which expression gets evaluated first? `(take 1000)`

isn’t even a valid call–where’s its second argument? How are *any* of these
forms evaluated?

What kind of arcane function *is* `->>`

?

All these mysteries, and more, in Chapter 5: Macros.

## Problems

- Write a function to find out if a string is a palindrome–that is, if it looks the same forwards and backwards.
- Find the number of ‘c’s in “abracadabra”.
- Write your own version of
`filter`

. - Find the first 100 prime numbers: 2, 3, 5, 7, 11, 13, 17, ….

Thank you! This is a great resource. Please keep it going. Let us know how we can help support you.

Are you sure the implementation of inc-more is correct? I believe it should call itself recursively instead of inc-first.

Thanks Greg, that is a typo. Fixed now. :)

Thank you for posting this.

I am looking into Lisp based languages these days, and as such I find it always good to get more basic information on language semantics.

There are a couple of things however that make a transition to Clojure very hard, and I think it would do some good if more information about these obstacles could be written down:

1) A lot of work needs to be invested to get a IDE going. I wasted a lot of time with emacs and Sublime Text 2/SublimeREPL before settling for Light Table. Note: I find the first two choices to be excellent editors, it is just that it is hard to get them going the way one wants to.

2) The JVM is said to be one of Clojures greatest assets. For me - I started programming with Python - reading java lingo error tracebacks and browsing the source code of JVM packages is very unpleasent experience.

An introduction on how to tackle the JVM library ‘problem’ in an efficient way a la Python would be great.

3) Compilation to JAR, running a service on GAE/Amazon without loosing much time on JVM startup and other practical things might be a nice to have in your guide.

4) Since multithreading is one of the key characteristics that sets Clojure apart from other lisp/scheme based languages e.g. the easy but great Chicken Scheme, it would be great to have a good look on parallel computing in the tutorial.

Cheers!

Thank you for posting this!!! This is the first tutorial I’ve done on Clojure that I feel has actually gotten me comfortable with the language. Can’t wait for the next installment :)

This is a great series. Content and delivery are both excellent! I have one question on the section about the infseq lazy sequence. How come that couldn’t simply be expressed as (take 10 (iterate inc 0))?

As a beginner in Clojure (and LISP in general), I could say that this is the most inspiring tutorial I know of for now! Looking forward for the next article.

Just demonstrating that lazy operations can be chained together. We’ll talk more about execution order and side effects in a later chapter. :)

Wow, what an expressive and powerful language Clojure is! The last code example really nailed it for me. Thanks for these series, as others already pointed out, they’re the best on the net. Keep them coming please…

Thank you for this Clojure series The exercises are also a great help as they do test one’s learnings

Hey, great tutorial! I’ve been programming Clojure for a little while, but I’m always looking for resources I can point people to who want to learn, and this is perfect. You also mentioned a few neat functions I didn’t know about (split-with, map-indexed, reductions, and reduced).

If you’ll pardon my nitpicking, I believe the solution to the problem under “Putting it all together” is incorrect. You state the problem as relating to “consecutive pairs of the first 1000 odd integers”, but your code operates on “the first 1000 consecutive pairs of odd integers”. That is, you include the pair ‘(1999 2001), even though 2001 is the 1001st odd integer. This could be fixed either by changing the problem statement, or by moving “(take 1000)” before/inside the “(partition …)”, yielding the answer 1331333001.

I stumbled upon this tutorial via your superb talks on distributed systems. Now I’m learning Clojure and currently going through “The joy of Clojure” book.

IMHO,

`recur`

would fit so nicely within the recursion topic. It’s interesting because I never came across something similar in Java, Scala nor (of course) in JS (maybe it’s there somewhere, but it’s certainly not widely used).BTW, inspiring tutorial intro.

Well I am new to Clojure. I started with “Clojure Programming” - but this introduction is just what I was looking for. Suggested edit - “The type map relates keys to values.” perhaps could read “The type map relates a set of keys to list of values.”

Brilliant. Excellent series on introduction to clojure.

What’s more idiomatic:

(fn pair (second pair))

or

#(apply * %)

?

Hi Kyle

Thank you for these very nice series of article on Clojure. Could you give us “the” solution — or the more idiomatic one — to the problems you list at the end of this article? I started solving them, but I’m kind of stuck on the third :/ https://gist.github.com/falzm/9543903

Cheers,

m.

Marc: Your version is just fine for this stage of the book. More idiomatically, you can use (mapcat …) in place of (apply concat (map …)). Probably the

cleanestsolution given the tools we’ve explored so far in the book would be`(defn filter [f coll] (reduce (fn [results x] (if (f x) (conj results x) results)) coll))`

But if I were writing this in idiomatic Clojure, I’d use a recursive lazy-seq, which we haven’t covered yet. ;-)

Thank you for this great tutorial.

Hi Kyle

I think your solution given above might be missing a default value for

`results`

.`(defn my-filter [f coll] (reduce (fn [results x] (if (f x) (conj results x) results)) [] coll))`

And, just forgot to say. Thanks for this series of article on Clojure! Learned a lot from them.

This is my favorite intro to recursion example I’ve seen so far!

these articles about clojure are awesome, most clojure book is hard to study for beginners of clojure, help me a lot.

great!

Great Work. I can tell you it’s not easy to find the clear explanations for a newbie. Most people lose track of what you need to know first when you don’t know. Their stuff is good once your going but it’s a rare teacher who can GET you going so you start to intuit how programming goes. Many THX

->> (Kyle) (Awesome)

Hey, I was wondering about tail optimization since Clojure is based on the JVM. So I tried the “transform-all” example on a very big vector and as excepted I got a StackOverflowError. I’ve since read that “recur” is aimed to circumvent this but I’m still learning about it so I can’t explain any further :) Maybe that’s something that could interest your readers (or it is mentionned later in the tutorial and haven’t read it yet)

Un gros merci pour ce beau tutoriel ! Jo

This is how tutorials should be. Precise and crisp! Thank you! This is my official “learn clojure in 3 days” guide :)

Hi guys,

Here are my solutions to the problems. Hope some of you find them helpful.

1.

(defn palindrome? [x]

`(= (seq x) (reverse (seq x)) )`

)

2.

( (frequencies (seq “abracadabra”)) \c )

3.

(defn my-filter [f x] (reduce (fn [output element]

`(if (f element) (conj output element) output ) ) [] x )`

)

4.

(defn isprime? n) (range 2 n)))))

(take 10 (filter isprime? (iterate inc 0)))

Hi Kyle,

Thanks for the amazing clojure tutorial; I’m really enjoying it so far!

One question about the complex problem at the end of this one: I’m pretty sure that in your solution, you’re taking the first 1000 pairs of odd integers rather than pairing the first 1000 odd integers. From my understanding, the solution to the problem as stated should read:

`(reduce + (map (fn [pair] (* (first pair) (second pair))) (partition 2 1 (take 1000 (filter odd? (iterate inc 0))))))`

I also want to say how much I appreciate these tutorials, this sequence-page alone has given me a lot of insight into how Clojure works. (In fact it’s the only page I’ve read so far, it came up when I was googling sequences - I definitely look forward to the rest of the articles).

I thought I’d post my solution to writing a filter also - not because it’s especially

goodcode, but because it’s bad - or at least veryprocedural.`(defn my_filter [func coll] (loop [output [] new_coll coll] (if (empty? new_coll) (println "output is " output) (if (func (first new_coll)) (recur (conj output (first new_coll)) (rest new_coll)) (recur output (rest new_coll))))))`

Check for palindrome:

(defn Palin? s))))

Returns true if the word is a palindrome, otherwise false.

Lovin' the tutorial. Coming from over 20 years of C style syntax this is really easing the learning curve. Thanks!

`(defn Palin? [s] (zero? (compare s (apply str (reverse s)))))`

Sorry for making such a mess above. I failed to read the fine print before posting and then didn’t follow through on the first attempt.

Still lovin' the tutorial…

Hey Not Chuck, why not just do this?

(defn Palin? s)))

;)

(sorry for the mess above) This is the correct piece of code to simplify your function:

`(defn palin? "returns true if its a palindrome, otherwise false" [x] (= x (apply str (reverse x))))`

Hey Kyle, loving this series so far. Any feedback on my version of filter?

`(defn my-filter "takes a predicate and a collection, applies the predicate to each item in the collection, returns a collection of all items that return true." [predicate collection] (if-let [item (first collection)] (if-let [true-item (predicate item)] (cons item (my-filter predicate (rest collection))) (my-filter predicate (rest collection))) collection))`

^ just realized i wasn’t using

`true-item`

and changed that`if-let`

to an`if`

Looks good, Chad. :)

Thank you so much for sharing this tutorial.

Awesome guide! Thanks!

Thanks again for the nice tutorial.

I want to suggest an improvement that is purely aesthetic (subject to personal taste).

I suggest replacing the sequence in this expression:

`(take 10 (iterate (fn [x] (if (odd? x) (+ 1 x) (/ x 2))) 10))`

by the much more interesting Collatz sequence:`(take 10 (iterate (fn [x] (if (odd? x) (+ 1 (* 3 x)) (/ x 2))) 10))`

We don’t know if this sequence ends in

`...2 1 2 1 2 1`

for each initial value eventually, whereas for the first sequence that is pretty easy to prove.Hi, very nice tutorial!! Thanks for very clear approach in the right pace.