在python中搜索一个以上字符串的列表

Question

I wrote the below function to return True if it detects the string 'Device' in a list

def data_filter2(inner_list):
   return any(['Device' in str(x) for x in inner_list])

is there a way to search the list for more than one string and return True it it finds either one?

I tried

def data_filter2(inner_list):
   return any(['Device' or 'Drug' in str(x) for x in inner_list])

But this did not work.

Answer 1

You could use a binary operator or or and depending on what you intend, but slightly different from how you've done it:

def data_filter2(inner_list):
   return any('Device' in str(x) or 'Drug' in str(x) for x in inner_list)

You could use a generator expression, and avoid creating a new list for your check.

Answer 2

What if you reverse the logic?

def data_filter2(inner_list):
   return any([str(x) in ['Device', 'Drug'] for x in inner_list])

This provides a framework that can "accept" more items to check for. Chaining or is not very pythonic to my eyes.

What is interesting to note here is the following:

alist_ = ['drugs', '123123']
astring = 'drug'


for i in range(len(alist_)):
    print(astring in alist_[i])
print('-----')

print(astring in alist_)

#prints:
#True
#False
#-----
#False

What this says is that searching for a string in a list requires the strings to be identical (case sensitive) but searching for a string in a string allows for substrings to return True as well.

Answer 3

The solutions using any and comprenhensions are nice. Another alternative would be to use set intersection.

In [30]: bool(set(['aaa', 'foo']) & set(['foo']))
Out[30]: True

In [31]: bool(set(['aaa', 'foo']) & set(['bar']))
Out[31]: False

Answer 4

You can do it in one line like so:

def data_filter2(inner_list):
   return any([x in y for x in ['Device', 'Drug'] for y in inner_list])