Counting substrings in a list of strings

Question

For example:

my_list = ["item*1", "item*2", "item*3", "item*4"]

I'd like a way to search for "item" and return 4 as item appears 4 times regardless of the extra characters.

You must iterate through the list and search in each string item for `"item"`. — Michael Butscher, Jun 11 '20 at 11:09
use `str.find()` method while iterating the strings in the list. or the `in` operator. — Adam, Jun 11 '20 at 11:10
what if there's `itemitemitem`? would you want that to return 1 or 3? — Nicolas Gervais, Jun 11 '20 at 11:36

score 5 · Answer 1 · answered Jun 11 '20 at 11:16

5

The simplest way I could come up with is this:

sum(map(lambda x: 'item' in x, my_list))

Out[1]: 4

It basically sums the True returned every time item is found in an element of my_list

answered Jun 11 '20 at 11:16

Nicolas Gervais

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But this one does not find multiple occurrences – Serafim Jun 11 '20 at 11:47
1

Since posting my answer I asked OP what behavior he expects for multiple occurrences. He has yet to answer. – Nicolas Gervais Jun 11 '20 at 11:48

LevB · Answer 2 · 2020-06-12T06:43:08.140

2

You can use count, checking each element in your list

my_list = ["item*1", "item*2", "item*3", "item*4"]

a = sum([el.count('item') for el in my_list])
print(a)

Update:

If you want "item" counted only once if it appears multiple times within elements of your list, the solution is even simpler.

a = sum(['item' in el for el in my_list])

edited Jun 12 '20 at 06:43

answered Jun 11 '20 at 11:17

LevB

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that will not return the right number if `item` is present twice in the same element – Nicolas Gervais Jun 11 '20 at 11:20
Yes, it would. `count`counts the number of occurrencies – Serafim Jun 11 '20 at 11:35
yes, that's what it does – Nicolas Gervais Jun 11 '20 at 11:37

score 1 · Answer 3 · answered Jun 11 '20 at 11:14

you can solve this in 2 ways. the first with the 'in operator and the second with the str.find() string method. and if your search is case insensetive use the str.lower() string class method.

count = 0
for word in my_list:
    if 'item' in word.lower():
        count +=1

print(count)

or using the str.find() method but the in is prefered

 count = 0
for word in my_list:
    if word.lower().find('item') != -1:
        count +=1

print(count)

score 1 · Answer 4 · answered Jun 11 '20 at 11:53

I've made a lambda function which will return the no of times the "item" appeared in a list or sequence.

count = lambda li: sum([i.count('item') for i in li])
print(count(["item*1", "item*2", "item*3", "item*4"]))

Saad · Answer 5 · 2020-06-13T23:22:09.160

1

You can convert the list into str and then use count method on the string to count the occurrence of the desired word in this case "item"

my_list = ["item*1", "item*2", "item*3", "item*4", 'itemitemitem']
count = ' '.join(my_list).count('item')
print(count)

This will work even if any "item" occurs more than one time like "itemitemitem"

edited Jun 13 '20 at 23:22

answered Jun 11 '20 at 11:59

Saad

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score 0 · Answer 6 · answered Jun 11 '20 at 11:11

0

data_is = ["item*1", "item*2", "item*3", "item*4"]
count = 0
for each in data_is:
  if 'item' in each:
    count = count +1

print(count)

answered Jun 11 '20 at 11:11

developer101

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7

score 0 · Answer 7 · answered Jun 11 '20 at 11:29

my_list = ["item*1", "item*2", "item*3", "item*4", "item-oh-item-oh-item"]
n = 0
pattern = "item"
for str in my_list:
    n += str.count(pattern)
print(n)

I added an extra string to show why I use count. If the pattern could be overlapping, like searching for "coco" in "cococo", I would use regular expressions from the package regex, not the standard package re.

Counting substrings in a list of strings

7 Answers7