Python: if list contains string print all the indexes / elements in the list that contain it

Question

I am able to detect matches but unable to locate where are they.

Given the following list:

['A second goldfish is nice and all', 3456, 'test nice']

I need to search for match (i.e. "nice") and print all the list elements that contain it. Ideally if the keyword to search were "nice" the results should be:

'A second goldfish is nice and all'
'test nice'

I have:

list = data_array
string = str(raw_input("Search keyword: "))
print string
if any(string in s for s in list):
    print "Yes"

So it finds the match and prints both, the keyword and "Yes" but it doesn't tell me where it is.

Should I iterate through every index in list and for each iteration search "string in s" or there is an easier way to do this?

Solution 1

Try this:

list = data_array
string = str(raw_input("Search keyword: "))
print string
for s in list:
    if string in str(s):
        print 'Yes'
        print list.index(s)

Editted to working example. If you only want the first matching index you can also break after the if statement evaluates true

Solution 2

matches = [s for s in my_list if my_string in str(s)]

or

matches = filter(lambda s: my_string in str(s), my_list)

Note that 'nice' in 3456 will raise a TypeError, which is why I used str() on the list elements. Whether that's appropriate depends on if you want to consider '45' to be in 3456 or not.

Solution 3

print filter(lambda s: k in str(s), l)

Solution 4

To print all the elements that contains nice

mylist = ['nice1', 'def456', 'ghi789', 'nice2', 'nice3']
sub = 'nice'
print("\n".join([e for e in mylist if sub in e]))
>>> nice1
    nice2
    nice3

To get the index of elements that contain nice (irrespective of the letter case)

mylist = ['nice1', 'def456', 'ghi789', 'Nice2', 'NicE3']
sub = 'nice'
index_list = []
i = 0
for e in mylist:
    if sub in e.lower():
        index_list.append(i)
    i +=1
print(index_list)
>>> [0, 3, 4]

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