Find all index position in list based on partial string inside item in list
Question:
mylist = ["aa123", "bb2322", "aa354", "cc332", "ab334", "333aa"]
I need the index position of all items that contain 'aa'. I’m having trouble combining enumerate() with partial string matching. I’m not even sure if I should be using enumerate.
I just need to return the index positions: 0,2,5
Answers:
Your idea to use enumerate() was correct.
indices = []
for i, elem in enumerate(mylist):
if 'aa' in elem:
indices.append(i)
Alternatively, as a list comprehension:
indices = [i for i, elem in enumerate(mylist) if 'aa' in elem]
You can use enumerate inside a list-comprehension:
indices = [i for i, s in enumerate(mylist) if 'aa' in s]
Without enumerate():
>>> mylist = ["aa123", "bb2322", "aa354", "cc332", "ab334", "333aa"]
>>> l = [mylist.index(i) for i in mylist if 'aa' in i]
>>> l
[0, 2, 5]
spell_list = ["Tuesday", "Wednesday", "February", "November", "Annual", "Calendar", "Solstice"]
index=spell_list.index("Annual")
print(index)
Based on this answer, I’d like to show how to "early exit" the iteration once the first item containing the substring aa is encountered. This only returns the first position.
import itertools
first_idx = len(tuple(itertools.takewhile(lambda x: "aa" not in x, mylist)))
This should be much more performant than looping over the whole list when the list is large, since takewhile will stop once the condition is False for the first time.
I know that the question asked for all positions, but since there will be many users stumbling upon this question when searching for the first substring, I’ll add this answer anyways.
mylist = ["aa123", "bb2322", "aa354", "cc332", "ab334", "333aa"]
I need the index position of all items that contain 'aa'. I’m having trouble combining enumerate() with partial string matching. I’m not even sure if I should be using enumerate.
I just need to return the index positions: 0,2,5
Your idea to use enumerate() was correct.
indices = []
for i, elem in enumerate(mylist):
if 'aa' in elem:
indices.append(i)
Alternatively, as a list comprehension:
indices = [i for i, elem in enumerate(mylist) if 'aa' in elem]
You can use enumerate inside a list-comprehension:
indices = [i for i, s in enumerate(mylist) if 'aa' in s]
Without enumerate():
>>> mylist = ["aa123", "bb2322", "aa354", "cc332", "ab334", "333aa"]
>>> l = [mylist.index(i) for i in mylist if 'aa' in i]
>>> l
[0, 2, 5]
spell_list = ["Tuesday", "Wednesday", "February", "November", "Annual", "Calendar", "Solstice"]
index=spell_list.index("Annual")
print(index)
Based on this answer, I’d like to show how to "early exit" the iteration once the first item containing the substring aa is encountered. This only returns the first position.
import itertools
first_idx = len(tuple(itertools.takewhile(lambda x: "aa" not in x, mylist)))
This should be much more performant than looping over the whole list when the list is large, since takewhile will stop once the condition is False for the first time.
I know that the question asked for all positions, but since there will be many users stumbling upon this question when searching for the first substring, I’ll add this answer anyways.