Check if a Python list item contains a string inside another string

Question

I have a list:

my_list = [\'abc-123\', \'def-456\', \'ghi-789\', \'abc-456\']

and want to search for items that contain the string \'

Answer 1

for item in my_list:
    if item.find("abc") != -1:
        print item

Answer 2

If you want to get list of data for multiple substrings

you can change it this way

some_list = ['abc-123', 'def-456', 'ghi-789', 'abc-456']
# select element where "abc" or "ghi" is included
find_1 = "abc"
find_2 = "ghi"
result = [element for element in some_list if find_1 in element or find_2 in element] 
# Output ['abc-123', 'ghi-789', 'abc-456']

Answer 3

Use filter to get at the elements that have abc.

>>> lst = ['abc-123', 'def-456', 'ghi-789', 'abc-456']
>>> print filter(lambda x: 'abc' in x, lst)
['abc-123', 'abc-456']

You can also use a list comprehension.

>>> [x for x in lst if 'abc' in x]

By the way, don't use the word list as a variable name since it is already used for the list type.

Answer 4

If you just need to know if 'abc' is in one of the items, this is the shortest way:

if 'abc' in str(my_list):

Answer 5

any('abc' in item for item in mylist)

Answer 6

From my knowledge, a 'for' statement will always consume time.

When the list length is growing up, the execution time will also grow.

I think that, searching a substring in a string with 'is' statement is a bit faster.

In [1]: t = ["abc_%s" % number for number in range(10000)]

In [2]: %timeit any("9999" in string for string in t)
1000 loops, best of 3: 420 µs per loop

In [3]: %timeit "9999" in ",".join(t)
10000 loops, best of 3: 103 µs per loop

But, I agree that the any statement is more readable.