I have two list:
main_list = ['Smith', 'Smith', 'Roger', 'Roger-Smith', '42']
master_list = ['Smith', 'Roger']
I want to count the number of times I find a string from master_list in a string of main_list without counting two times the same item.
Example: for the two lists above, the result of my function should be 4. 'Smith' can be retrieved 3 times in main_list. 'Roger can be found 2 times but as 'Smith' was already found in 'Roger-Smith', this one doesn't count anymore, so 'Roger' is just count as 1 which make 4 in total.
The function I wrote for know is below but I think there is a faster way to do it:
def string_detection(master_list, main_list):
count = 0
for substring in master_list:
temp = list(main_list)
for string in temp:
if substring in string:
main_list.remove(string)
count+=1
return count
A one liner
>>>sum(any(m in L for m in master_list) for L in main_list)
4
Iterate over main_list
and check if any
of the values from master_list
are in that string. This leaves you with a list of bool values. It will stop after it finds one and so adds only one to the count for each string. Conveniently sum
counts all the True
s to give you the count.
You can use pandas
(which provide fast vectorized operations) with str.contains
and sum()
import pandas as pd
main_list = pd.Series(['Smith', 'Smith', 'Roger', 'Roger-Smith', '42'])
master_list = ['Smith', 'Roger']
count = main_list.str.contains('|'.join(master_list)).sum()
You can do it other way around. Create list that will contain only elements from main_list
that have substring from master_list
temp_list = [ string for string in main_list if any(substring in string for substring in master_list)]
Now temp_list
looks like this:
['Smith', 'Smith', 'Roger', 'Roger-Smith']
So the length of temp_list
is your answer.
What about this
main_list = ['Smith', 'Smith', 'Roger', 'Roger-Smith', '42']
master_list = ['Smith', 'Roger']
print len([word for word in main_list if any(mw in word for mw in master_list)])
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