如何检查字符串是否是字符串列表中项目的子字符串？

Question

How do I search for items that contain the string 'abc' in the following list?

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

The following checks if 'abc' is in the list, but does not detect 'abc-123' and 'abc-456' :

if 'abc' in xs:

Answer 1

To check for the presence of 'abc' in any string in the list:

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

if any("abc" in s for s in xs):
    ...

---

To get all the items containing 'abc' :

matching = [s for s in xs if "abc" in s]

Answer 2

Just throwing this out there: if you happen to need to match against more than one string, for example abc and def , you can combine two comprehensions as follows:

matchers = ['abc','def']
matching = [s for s in my_list if any(xs in s for xs in matchers)]

Output:

['abc-123', 'def-456', 'abc-456']

Answer 3

Use filter to get all the elements that have 'abc' :

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

One can also use a list comprehension:

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

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):

Note: this assumes 'abc' is an alphanumeric text. Do not use it if 'abc' could be just a special character (ie []', ).

Answer 5

This is quite an old question, but I offer this answer because the previous answers do not cope with items in the list that are not strings (or some kind of iterable object). Such items would cause the entire list comprehension to fail with an exception.

To gracefully deal with such items in the list by skipping the non-iterable items, use the following:

[el for el in lst if isinstance(el, collections.Iterable) and (st in el)]

then, with such a list:

lst = [None, 'abc-123', 'def-456', 'ghi-789', 'abc-456', 123]
st = 'abc'

you will still get the matching items ( ['abc-123', 'abc-456'] )

The test for iterable may not be the best. Got it from here: In Python, how do I determine if an object is iterable?

Answer 6

x = 'aaa'
L = ['aaa-12', 'bbbaaa', 'cccaa']
res = [y for y in L if x in y]

Answer 7

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

Answer 8

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

Answer 9

I am new to Python. I got the code below working and made it easy to understand:

my_list = ['abc-123', 'def-456', 'ghi-789', 'abc-456']
for item in my_list:
    if 'abc' in item:
       print(item)

Answer 10

Use the __contains__() method of Pythons string class.:

a = ['abc-123', 'def-456', 'ghi-789', 'abc-456']
for i in a:
    if i.__contains__("abc") :
        print(i, " is containing")

Answer 11

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 12

I needed the list indices that correspond to a match as follows:

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

[n for n, x in enumerate(lst) if 'abc' in x]

output

[0, 3]

Answer 13

mylist=['abc','def','ghi','abc']

pattern=re.compile(r'abc') 

pattern.findall(mylist)

Answer 14

Adding nan to list, and the below works for me:

some_list = ['abc-123', 'def-456', 'ghi-789', 'abc-456',np.nan]
any([i for i in [x for x in some_list if str(x) != 'nan'] if "abc" in i])

Answer 15

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

for item in my_list:
    if (item.find('abc')) != -1:
        print ('Found at ', item)

Answer 16

I did a search, which requires you to input a certain value, then it will look for a value from the list which contains your input:

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

imp = raw_input('Search item: ')

for items in my_list:
    val = items
    if any(imp in val for items in my_list):
        print(items)

Try searching for 'abc'.

Answer 17

def find_dog(new_ls):
    splt = new_ls.split()
    if 'dog' in splt:
        print("True")
    else:
        print('False')


find_dog("Is there a dog here?")

Answer 18

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.

Answer 19

Question : Give the informations of abc

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


    aa = [ string for string in a if  "abc" in string]
    print(aa)

Output =>  ['abc-123', 'abc-456']