I'm trying to create a function:
filter(delete,lst)
When someone inputs:
filter(1,[1,2,1])
returns [2]
What I have come up with was to use the list.remove function but it only deletes the first instance of delete.
def filter(delete, lst):
"""
Removes the value or string of delete from the list lst
"""
list(lst)
lst.remove(delete)
print lst
My result:
filter(1,[1,2,1])
returns [2,1]
Try with list comprehensions:
def filt(delete, lst):
return [x for x in lst if x != delete]
Or alternatively, with the built-in filter function:
def filt(delete, lst):
return filter(lambda x: x != delete, lst)
And it's better not to use the name filter for your function, since that's the same name as the built-in function used above
filter as a name, since it's already a built-in.
I like the answer from Óscar López but you should also learn to use the existing Python filter function:
>>> def myfilter(tgt, seq):
return filter(lambda x: x!=tgt, seq)
>>> myfilter(1, [1,2,1])
[2]
def my_filter(func,sequence):
res=[]
for variable in sequence :
if func(variable):
res.append(variable)
return res
def is_even(item):
if item%2==0 :
return True
else :
return False
seq=[1,2,3,4,5,6,7,8,9,10]
print(my_filter(is_even,seq))
Custom filter function in python:
def myfilter(fun, data):
return [i for i in data if fun(i)]
filterfunction and you'll be masking it.