如何获取项目在列表中的位置？

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

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.

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

Use enumerate:

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

for i in xrange(len(testlist)):
  if testlist[i] == 1:
    print i

xrange instead of range as requested (see comments).

Here is another way to do this:

try:
   id = testlist.index('1')
   print testlist[id]
except ValueError:
   print "Not Found"

[x for x in range(len(testlist)) if testlist[x]==1]

Try the below:

testlist = [1,2,3,5,3,1,2,1,6]    
position=0
for i in testlist:
   if i == 1:
      print(position)
   position=position+1

If your list got large enough and you only expected to find the value in a sparse number of indices, consider that this code could execute much faster because you don’t have to iterate every value in the list.

lookingFor = 1
i = 0
index = 0
try:
  while i < len(testlist):
    index = testlist.index(lookingFor,i)
    i = index + 1
    print index
except ValueError: #testlist.index() cannot find lookingFor
  pass

If you expect to find the value a lot you should probably just append “index” to a list and print the list at the end to save time per iteration.

I think that it might be useful to use the curselection() method from thte Tkinter library:

from Tkinter import * 
listbox.curselection()

This method works on Tkinter listbox widgets, so you’ll need to construct one of them instead of a list.

This will return a position like this:

(‘0’,) (although later versions of Tkinter may return a list of ints instead)

Which is for the first position and the number will change according to the item position.

For more information, see this page: http://effbot.org/tkinterbook/listbox.htm

Greetings.

Why complicate things?

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

Just to illustrate complete example along with the input_list which has searies1 (example: input_list[0]) in which you want to do a lookup of series2 (example: input_list[1]) and get indexes of series2 if it exists in series1.

Note: Your certain condition will go in lambda expression if conditions are simple

input_list = [[1,2,3,4,5,6,7],[1,3,7]]
series1 = input_list[0]
series2 = input_list[1]
idx_list = list(map(lambda item: series1.index(item) if item in series1 else None, series2))
print(idx_list)

output:

[0, 2, 6]

testlist = [1,2,3,5,3,1,2,1,6]
for id, value in enumerate(testlist):
    if id == 1:
        print testlist[id]

I guess that it’s exacly what you want. ;-) ‘id’ will be always the index of the values on the list.