How do i move the offset of the 'index' method of 'list'

Question

i know that the index function work like this:

list = ['dog','cat','pizza','trump', 'computer', 'trump']
print list.index('trump')

the output will be 3. but now i want him for print the other 'trump' string, that come after 2 objects. but if i will do the same command:

print list.index('trump')

he will print 3 again - the first trump he see. so how do i move the 'offset' of the index function, so she will detect the other trump, in the index 5? thanks lot you guys!!

are you looking to print all the indexes where your string occur? — mugabits, May 23 '17 at 14:25
You might find this useful https://stackoverflow.com/questions/6294179/how-to-find-all-occurrences-of-an-element-in-a-list — carthurs, May 23 '17 at 14:25
BTW, don't use the variable name `list` as it shadows the built-in function `list()` — Chris_Rands, May 23 '17 at 14:29

score 5 · Answer 1 · answered May 23 '17 at 14:26

list.index takes a 2nd start argument:

>>> lst = ['dog','cat','pizza','trump', 'computer', 'trump']
>>> lst.index('trump') # 1st index
3
>>> lst.index('trump',4) # 2nd index
5
>>> lst.index('trump',lst.index('trump')+1) # 2nd index
5

Or if you want all indexes, use a list comprehension:

>>> [idx for idx,item in enumerate(lst) if item == 'trump']
[3, 5]

Jean-François Fabre · Answer 2 · 2017-05-23T14:33:28.730

Just note the first index where the string appears, and pass the extra parameter to index the second time:

l = ['dog','cat','pizza','trump', 'computer', 'trump']
i = l.index("trump")
print(i)
print(l.index("trump",i+1))

I get:

3
5

from the inline doc, you can pass an optional start & stop value:

index(...)

L.index(value, [start, [stop]]) -> integer -- return first index of value. Raises ValueError if the value is not present.

(in a general case you have to protect your call to index by a try/except ValueError block in case the value is not in the list and act accordingly)

Note that start can be negative. If negative, the index is computed from the end of the list instead.

score 2 · Answer 3 · answered May 23 '17 at 14:26

2

If you want a list of all indices for an entry in your list, you can do something like this:

indices = [i for i, x in enumerate(list) if x == "key"]

answered May 23 '17 at 14:26

Christoph Engwer

781
6
9

score 1 · Answer 4 · answered May 23 '17 at 14:26

1

create a shortened version of the list.

list=['dog','cat','pizza','trump', 'computer', 'trump']
print list.index('trump')
list = list[list.index('trump')+1:]
print list.index('trump')

answered May 23 '17 at 14:26

E.D.

639
6
14

score 1 · Answer 5 · answered May 23 '17 at 14:31

1

l = ['dog','cat','pizza','trump', 'computer', 'trump']
i = [n for n, item in enumerate(l) if item == 'trump'] #find all indices

# print the result
for n in i:
   print n, l[n]

3 trump
5 trump

answered May 23 '17 at 14:31

alec_djinn

10,104
8
46
71

score 0 · Answer 6 · answered May 23 '17 at 19:55

Try this...it work like charm.

Actually in your problem iteration executes only once even in for loop it wont work(not work for me even after iter and next() used)so I tried with while then I realize it is the game of iteration so put while in try with while 1: to start with iteration 1and index method does a linear search, and stops at the first matching item. If no matching item is found, it raises a ValueError exception.

CODE

list = ['dog','cat','pizza','trump', 'computer', 'trump']
value = "trump"
i = -1
try:
    while 1: # Start with Iteration 1 never 0
        i = list.index(value, i+1)
        print(value,"at", i)
except ValueError: # No matching item is found.
    pass

OUTPUT

trump at 3
trump at 5

Found another easiest way to get common string in list #CODE `for idx,val in enumerate(l): if val=='trump': print(idx,val)` #OUTPUT `3 trump 5 trump` — Vrushal Raut, May 24 '17 at 07:50

How do i move the offset of the 'index' method of 'list'

6 Answers6

CODE

OUTPUT