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问题:
I found, that there is related question, about how to find if at least one item exists in a list:
How to check if one of the following items is in a list?
But what is the best and pythonic way to find whether all items exists in a list?
Searching through the docs I found this solution:
>>> l = ['a', 'b', 'c']
>>> set(['a', 'b']) <= set(l)
True
>>> set(['a', 'x']) <= set(l)
False
Other solution would be this:
>>> l = ['a', 'b', 'c']
>>> all(x in l for x in ['a', 'b'])
True
>>> all(x in l for x in ['a', 'x'])
False
But here you must do more typing.
Is there any other solutions?
回答1:
Operators like <=
in Python are generally not overriden to mean something significantly different than "less than or equal to". It's unusual for the standard library does this--it smells like legacy API to me.
Use the equivalent and more clearly-named method, set.issubset
. Note that you don't need to convert the argument to a set; it'll do that for you if needed.
set(['a', 'b']).issubset(['a', 'b', 'c'])
回答2:
I would probably use set
in the following manner :
set(l).issuperset(set(['a','b']))
or the other way round :
set(['a','b']).issubset(set(l))
I find it a bit more readable, but it may be over-kill. Sets are particularly useful to compute union/intersection/differences between collections, but it may not be the best option in this situation ...
回答3:
I like these two because they seem the most logical, the latter being shorter and probably fastest (shown here using set
literal syntax which has been backported to Python 2.7):
all(x in {'a', 'b', 'c'} for x in ['a', 'b'])
# or
{'a', 'b'}.issubset({'a', 'b', 'c'})
回答4:
What if your lists contain duplicates like this:
v1 = ['s', 'h', 'e', 'e', 'p']
v2 = ['s', 's', 'h']
Sets do not contain duplicates. So, the following line returns True.
set(v2).issubset(v1)
To count for duplicates, you can use the code:
v1 = sorted(v1)
v2 = sorted(v2)
def is_subseq(v2, v1):
"""Check whether v2 is a subsequence of v1."""
it = iter(v1)
return all(c in it for c in v2)
So, the following line returns False.
is_subseq(v2, v1)
回答5:
This is a different method:
Code:
def isAll(lis1, lis2):
list = []
for i in lis2:
for ii in lis1:
if i == ii:
list.append(True)
else:
list.append(False)
if list.count(True) == len(lis1):
return True
else:
return False
lis1 = ["a","b","c"]
lis2 = ["a","b"]
isAll(lis1, lis2) #Output: False