I want to check if a va开发者_JAVA技巧lue is in a list, no matter what the case of the letters are, and I need to do it efficiently.
This is what I have:
if val in list:
But I want it to ignore case
check = "asdf"
checkLower = check.lower()
print any(checkLower == val.lower() for val in ["qwert", "AsDf"])
# prints true
Using the any() function. This method is nice because you aren't recreating the list to have lowercase, it is iterating over the list, so once it finds a true value, it stops iterating and returns.
Demo : http://codepad.org/dH5DSGLP
If you know that your values are all of type str
or unicode
, you can try this:
if val in map(str.lower, list):
...Or:
if val in map(unicode.lower, list):
If you really have just a list of the values, the best you can do is something like
if val.lower() in [x.lower() for x in list]: ...
but it would probably be better to maintain, say, a set
or dict
whose keys are lowercase versions of the values in the list; that way you won't need to keep iterating over (potentially) the whole list.
Incidentally, using list
as a variable name is poor style, because list
is also the name of one of Python's built-in types. You're liable to find yourself trying to call the list
builtin function (which turns things into lists) and getting confused because your list
variable isn't callable. Or, conversely, trying to use your list
variable somewhere where it happens to be out of scope and getting confused because you can't index into the list
builtin.
You can lower the values and check them:
>>> val
'CaSe'
>>> l
['caSe', 'bar']
>>> val in l
False
>>> val.lower() in (i.lower() for i in l)
True
items = ['asdf', 'Asdf', 'asdF', 'asjdflk', 'asjdklflf']
itemset = set(i.lower() for i in items)
val = 'ASDF'
if val.lower() in itemset: # O(1)
print('wherever you go, there you are')
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