Solving problem is about exposing yourself to as many situations as possible like What does enumerate() mean? and practice these strategies over and over. With time, it becomes second nature and a natural way you approach any problems in general. Big or small, always start with a plan, use other strategies mentioned here till you are confident and ready to code the solution.
In this post, my aim is to share an overview the topic about What does enumerate() mean?, which can be followed any time. Take easy to follow this discuss.
for row_number, row in enumerate(cursor): do in Python?
enumerate mean in this context?
enumerate() function adds a counter to an iterable.
So for each element in
cursor, a tuple is produced with
(counter, element); the
for loop binds that to
'foo', 'bar', 'baz') for elem in elements: print elem foo bar baz for count, elem in enumerate(elements): print count, elem 0 foo 1 bar 2 bazelements = (
enumerate() starts counting at
0 but if you give it a second integer argument, it’ll start from that number instead:
for count, elem in enumerate(elements, 42): print count, elem 42 foo 43 bar 44 baz
from itertools import count def enumerate(it, start=0): # return an iterator that adds a counter to each element of it return zip(count(start), it)
def enumerate(it, start=0): count = start for elem in it: yield (count, elem) count += 1
The actual implementation in C is closer to the latter, with optimisations to reuse a single tuple object for the common
for i, ... unpacking case and using a standard C integer value for the counter until the counter becomes too large to avoid using a Python integer object (which is unbounded).
It’s a builtin function that returns an object that can be iterated over. See the documentation.
In short, it loops over the elements of an iterable (like a list), as well as an index number, combined in a tuple:
for item in enumerate(["a", "b", "c"]): print item
(0, "a") (1, "b") (2, "c")
It’s helpful if you want to loop over a sequence (or other iterable thing), and also want to have an index counter available. If you want the counter to start from some other value (usually 1), you can give that as second argument to
I am reading a book (Effective Python) by Brett Slatkin and he shows another way to iterate over a list and also know the index of the current item in the list but he suggests that it is better not to use it and to use
I know you asked what enumerate means, but when I understood the following, I also understood how
enumerate makes iterating over a list while knowing the index of the current item easier (and more readable).
list_of_letters = ['a', 'b', 'c'] for i in range(len(list_of_letters)): letter = list_of_letters[i] print (i, letter)
The output is:
0 a 1 b 2 c
I also used to do something, even sillier before I read about the
i = 0 for n in list_of_letters: print (i, n) i += 1
It produces the same output.
enumerate I just have to write:
list_of_letters = ['a', 'b', 'c'] for i, letter in enumerate(list_of_letters): print (i, letter)
As other users have mentioned,
enumerate is a generator that adds an incremental index next to each item of an iterable.
So if you have a list say
l = ["test_1", "test_2", "test_3"], the
list(enumerate(l)) will give you something like this:
[(0, 'test_1'), (1, 'test_2'), (2, 'test_3')].
Now, when this is useful? A possible use case is when you want to iterate over items, and you want to skip a specific item that you only know its index in the list but not its value (because its value is not known at the time).
for index, value in enumerate(joint_values): if index == 3: continue # Do something with the other `value`
So your code reads better because you could also do a regular for loop with
range but then to access the items you need to index them (i.e.,
Although another user mentioned an implementation of
zip, I think a more pure (but slightly more complex) way without using
itertools is the following:
def enumerate(l, start=0): return zip(range(start, len(l) + start), l)
l = ["test_1", "test_2", "test_3"] enumerate(l) enumerate(l, 10)
[(0, ‘test_1’), (1, ‘test_2’), (2, ‘test_3’)]
[(10, ‘test_1’), (11, ‘test_2’), (12, ‘test_3’)]
As mentioned in the comments, this approach with range will not work with arbitrary iterables as the original
enumerate function does.
The enumerate function works as follows:
doc = """I like movie. But I don't like the cast. The story is very nice""" doc1 = doc.split('.') for i in enumerate(doc1): print(i)
The output is
(0, 'I like movie') (1, " But I don't like the cast") (2, ' The story is very nice')
I am assuming that you know how to iterate over elements in some list:
for el in my_list: # do something
Now sometimes not only you need to iterate over the elements, but also you need the index for each iteration. One way to do it is:
i = 0 for el in my_list: # do somethings, and use value of "i" somehow i += 1
However, a nicer way is to user the function “enumerate”. What enumerate does is that it receives a list, and it returns a list-like object (an iterable that you can iterate over) but each element of this new list itself contains 2 elements: the index and the value from that original input list:
So if you have
arr = ['a', 'b', 'c']
Then the command
returns something like:
[(0,'a'), (1,'b'), (2,'c')]
Now If you iterate over a list (or an iterable) where each element itself has 2 sub-elements, you can capture both of those sub-elements in the
for loop like below:
for index, value in enumerate(arr): print(index,value)
which would print out the sub-elements of the output of enumerate.
And in general you can basically “unpack” multiple items from list into multiple variables like below:
idx,value = (2,'c') print(idx) print(value)
which would print
This is the kind of assignment happening in each iteration of that loop with
enumerate(arr) as iterable.