Python enumerate list setting start index but without increasing end count

Question

I want to loop through a list with a counter starting at zero but with the list starting index at 1 eg:

valueList = [1, 2, 3, 4]
secondList = ['a', 'b', 'c', 'd']

for i, item in enumerate(valueList, start=1):
    print(secondList[i])

The code fails with an index out of range error (I realize this is because i ends at the length of the list -1 plus the start value and that python lists are zero indexed so using i in this way to call the ith element in another list is not valid). The following works but the addition of the test for i greater than zero looks un-pythonic.

valueList = [1, 2, 3, 4]
secondList = ['a', 'b', 'c', 'd']

for i, item in enumerate(valueList, start=0):
    if i > 0:  
        print(secondList[i]) 

Enumerate is not the right choice, is there another way?

Answer 1

It sounds as if you want to slice the list instead; still start enumerate() at one to get the same indices:

for i, item in enumerate(valueList[1:], start=1):

This then loops over valueList starting at the second element, with matching indices:

>>> valueList = [1, 2, 3, 4]
>>> secondList = ['a', 'b', 'c', 'd']
>>> for i, item in enumerate(valueList[1:], start=1):
...     print(secondList[i])
... 
b
c
d

In this case, I'd just use zip() instead, perhaps combined with itertools.islice() :

from itertools import islice

for value, second in islice(zip(valueList, secondList), 1, None):
    print(value, second)

The islice() call skips the first element for you:

>>> from itertools import islice
>>> for value, second in islice(zip(valueList, secondList), 1, None):
...     print(value, second)
... 
2 b
3 c
4 d

Answer 2

The issue is not enumerate, and neither the start argument, but the fact that when you do start=1 , you're enumerating from 1 to valueList+1 :

>>> valueList = [1, 2, 3, 4]
>>> secondList = ['a', 'b', 'c', 'd']
>>> for i, item in enumerate(valueList, start=1):
...     print(i)
...     print(secondList[i])
...     print('----')
... 
1
b
----
2
c
----
3
d
----
4
Traceback (most recent call last):
  File "<stdin>", line 3, in <module>
IndexError: list index out of range

So of course, when you try to access secondList[4] there's no value available! You might want to do:

>>> for i, item in enumerate(valueList, start=1):
...     if i < len(secondList):
...         print(secondList[i])
... 
b
c
d

That said, I'm not sure what you're exactly attempting to achieve. If you want to skip the first value of secondList , that might be a solution, even though not the most efficient one. A better way would be to actually use the slice operator:

>>> print(secondList[1:])
['b', 'c', 'd']

If you want to iterate over a list using natural enumeration (instead of computer 's one), ie starting from 1 instead of 0 , then that's not the way to go. To show natural indexes and use computer indexes, you just have to do:

>>> for i, item in enumerate(valueList):
...     print("{} {}".format(i+1, secondList[i]))
... 
1 a
2 b
3 c
4 d

Finally, you could use zip() instead of enumerate to link contents of both lists:

>>> for i, item in zip(valueList, secondList):
...     print('{} {}'.format(i, item))
... 
1 a
2 b
3 c
4 d

which will show each value of valueList attached with the value of secondList at the same index.

Answer 3

If your collections are generators, you might be best off doing:

for i, item in enumerate(valueList, start=1):
    if i < 1:
        continue
    print(secondList[i])

or to modify Martijn Pieters's great answer,

from itertools import islice, izip
for value, second in islice(izip(valueList, secondList), 1, None):
    print(value, second)

this way data are lazily loaded.

Answer 4

Or without using the start-parameter, you can do an if-comparison i > 0. That way you also don't have to deal with the list+1 index error at the end:

for i, item in enumerate(valueList):
    if i > 0:
        continue
    print(secondList[i])