How to get item's position in a list?

Question

I am iterating over a list and I want to print out the index of the item if it meets a certain condition. How would I do this?

Example:

testlist = [1,2,3,5,3,1,2,1,6] for item in testlist:     if item == 1:         print position 

Answer 1

Hmmm. There was an answer with a list comprehension here, but it's disappeared.

Here:

 [i for i,x in enumerate(testlist) if x == 1] 

Example:

>>> testlist [1, 2, 3, 5, 3, 1, 2, 1, 6] >>> [i for i,x in enumerate(testlist) if x == 1] [0, 5, 7] 

Update:

Okay, you want a generator expression, we'll have a generator expression. Here's the list comprehension again, in a for loop:

>>> for i in [i for i,x in enumerate(testlist) if x == 1]: ...     print i ...  0 5 7 

Now we'll construct a generator...

>>> (i for i,x in enumerate(testlist) if x == 1) <generator object at 0x6b508> >>> for i in (i for i,x in enumerate(testlist) if x == 1): ...     print i ...  0 5 7 

and niftily enough, we can assign that to a variable, and use it from there...

>>> gen = (i for i,x in enumerate(testlist) if x == 1) >>> for i in gen: print i ...  0 5 7 

And to think I used to write FORTRAN.

Answer 2

What about the following?

print testlist.index(element) 

If you are not sure whether the element to look for is actually in the list, you can add a preliminary check, like

if element in testlist:     print testlist.index(element) 

or

print(testlist.index(element) if element in testlist else None) 

or the "pythonic way", which I don't like so much because code is less clear, but sometimes is more efficient,

try:     print testlist.index(element) except ValueError:     pass