python - Is there a need for range(len(a))?

Question

One frequently finds expressions of this type in python questions on SO. Either for just accessing all items of the iterable

for i in range(len(a)):
    print(a[i])

Which is just a clumbersome way of writing:

for e in a:
    print(e)

Or for assigning to elements of the iterable:

for i in range(len(a)):
    a[i] = a[i] * 2

Which should be the same as:

for i, e in enumerate(a):
     a[i] = e * 2
# Or if it isn't too expensive to create a new iterable
a = [e * 2 for e in a]

Or for filtering over the indices:

for i in range(len(a)):
    if i % 2 == 1: continue
    print(a[i])

Which could be expressed like this:

for e in a [::2]:
    print(e)

Or when you just need the length of the list, and not its content:

for _ in range(len(a)):
    doSomethingUnrelatedToA()

Which could be:

for _ in a:
    doSomethingUnrelatedToA()

In python we have enumerate, slicing, filter, sorted, etc... As python for constructs are intended to iterate over iterables and not only ranges of integers, are there real-world use-cases where you need in range(len(a))?

Answer 1

If you need to work with indices of a sequence, then yes - you use it... eg for the equivalent of numpy.argsort...:

>>> a = [6, 3, 1, 2, 5, 4]
>>> sorted(range(len(a)), key=a.__getitem__)
[2, 3, 1, 5, 4, 0]

Answer 2

What if you need to access two elements of the list simultaneously?

for i in range(len(a[0:-1])):
    something_new[i] = a[i] * a[i+1]

You can use this, but it's probably less clear:

for i, _ in enumerate(a[0:-1]):
     something_new[i] = a[i] * a[i+1]

Personally I'm not 100% happy with either!

Answer 3

Short answer: mathematically speaking, no, in practical terms, yes, for example for Intentional Programming.

Technically, the answer would be "no, it's not needed" because it's expressible using other constructs. But in practice, I use for i in range(len(a) (or for _ in range(len(a)) if I don't need the index) to make it explicit that I want to iterate as many times as there are items in a sequence without needing to use the items in the sequence for anything.

So: "Is there a need?"? — yes, I need it to express the meaning/intent of the code for readability purposes.

See also: https://en.wikipedia.org/wiki/Intentional_programming

And obviously, if there is no collection that is associated with the iteration at all, for ... in range(len(N)) is the only option, so as to not resort to i = 0; while i < N; i += 1 ...

Answer 4

Going by the comments as well as personal experience, I say no, there is no need for range(len(a)). Everything you can do with range(len(a)) can be done in another (usually far more efficient) way.

You gave many examples in your post, so I won't repeat them here. Instead, I will give an example for those who say "What if I want just the length of a, not the items?". This is one of the only times you might consider using range(len(a)). However, even this can be done like so:

>>> a = [1, 2, 3, 4]
>>> for _ in a:
...     print True
...
True
True
True
True
>>>

Clements answer (as shown by Allik) can also be reworked to remove range(len(a)):

>>> a = [6, 3, 1, 2, 5, 4]
>>> sorted(range(len(a)), key=a.__getitem__)
[2, 3, 1, 5, 4, 0]
>>> # Note however that, in this case, range(len(a)) is more efficient.
>>> [x for x, _ in sorted(enumerate(a), key=lambda i: i[1])]
[2, 3, 1, 5, 4, 0]
>>>

So, in conclusion, range(len(a)) is not needed. Its only upside is readability (its intention is clear). But that is just preference and code style.

Answer 5

Sometimes matplotlib requires range(len(y)), e.g., while y=array([1,2,5,6]), plot(y) works fine, scatter(y) does not. One has to write scatter(range(len(y)),y). (Personally, I think this is a bug in scatter; plot and its friends scatter and stem should use the same calling sequences as much as possible.)

Answer 6

It's nice to have when you need to use the index for some kind of manipulation and having the current element doesn't suffice. Take for instance a binary tree that's stored in an array. If you have a method that asks you to return a list of tuples that contains each nodes direct children then you need the index.

#0 -> 1,2 : 1 -> 3,4 : 2 -> 5,6 : 3 -> 7,8 ...
nodes = [0,1,2,3,4,5,6,7,8,9,10]
children = []
for i in range(len(nodes)):
  leftNode = None
  rightNode = None
  if i*2 + 1 < len(nodes):
    leftNode = nodes[i*2 + 1]
  if i*2 + 2 < len(nodes):
    rightNode = nodes[i*2 + 2]
  children.append((leftNode,rightNode))
return children

Of course if the element you're working on is an object, you can just call a get children method. But yea, you only really need the index if you're doing some sort of manipulation.

Answer 7

I have an use case I don't believe any of your examples cover.

boxes = [b1, b2, b3]
items = [i1, i2, i3, i4, i5]
for j in range(len(boxes)):
    boxes[j].putitemin(items[j])

I'm relatively new to python though so happy to learn a more elegant approach.

Answer 8

If you have to iterate over the first len(a) items of an object b (that is larger than a), you should probably use range(len(a)):

for i in range(len(a)):
    do_something_with(b[i])

Answer 9

Sometimes, you really don't care about the collection itself. For instance, creating a simple model fit line to compare an "approximation" with the raw data:

fib_raw = [1, 1, 2, 3, 5, 8, 13, 21] # Fibonacci numbers

phi = (1 + sqrt(5)) / 2
phi2 = (1 - sqrt(5)) / 2

def fib_approx(n): return (phi**n - phi2**n) / sqrt(5)

x = range(len(data))
y = [fib_approx(n) for n in x]

# Now plot to compare fib_raw and y
# Compare error, etc

In this case, the values of the Fibonacci sequence itself were irrelevant. All we needed here was the size of the input sequence we were comparing with.

Answer 10

Very simple example:

def loadById(self, id):
    if id in range(len(self.itemList)):
        self.load(self.itemList[id])

I can't think of a solution that does not use the range-len composition quickly.

But probably instead this should be done with try .. except to stay pythonic i guess..

Answer 11

My code is:

s=["9"]*int(input())
for I in range(len(s)):
    while not set(s[I])<=set('01'):s[i]=input(i)
print(bin(sum([int(x,2)for x in s]))[2:])

It is a binary adder but I don't think the range len or the inside can be replaced to make it smaller/better.

Answer 12

Some people said that for i in range len(a) is for beginners,but what if I want to access to the element and the indices simultenously.