I have a list:
my_list = ['abc-123', 'def-456', 'ghi-789', 'abc-456']
and want to search for items that contain the string 'abc'. How can I do that?
if 'abc' in my_list:
would check if 'abc' exists in the list but it is a part of 'abc-123' and 'abc-456', 'abc' does not exist on its own. So how can I get all items that contain 'abc' ?
my_list = ['abc-123', 'def-456', 'ghi-789', 'abc-456']
for item in my_list:
if (item.find('abc')) != -1:
print ('Found at ', item)
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 a just a special character (i.e. []', ).
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'.
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']
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?")
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']
mylist=['abc','def','ghi','abc']
pattern=re.compile(r'abc')
pattern.findall(mylist)
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 str in my_list:
if 'abc' in str:
print(str)
any('abc' in item for item in mylist)
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']
x = 'aaa'
L = ['aaa-12', 'bbbaaa', 'cccaa']
res = [y for y in L if x in y]
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.
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]
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])
for item in my_list:
if item.find("abc") != -1:
print item
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?
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")
Source: Stackoverflow.com