Reputation: 43
So my intention is to iterate over lists in lists with map, filter and lambda e.g.:
[ [33,3,3,0] , [34,1,0,4] ] should result in [ [33,3,3] , [34,1,4] ]
def no_zeros(lst):
a=map(filter(lambda x:(y==0 for y in x) ,lst),lst)
print (list(a))
#(says that filter is not callable)
I know that this code won't work but i do not know what exactly i do wrong since i'm still kind of new to lambda,etc.I already tried different stuff , like moving the filter after the lambda,but that did not seem to work either . Im gratefull for any kind of tipps or help.
Upvotes: 4
Views: 8646
Reputation: 322
Another solution might be:
def no_zeros(lst):
if isinstance(lst, list):
for i in lst:
while 0 in i:
i.remove(0)
print(lst)
else:
raise TypeError("lst must be type of list")
Upvotes: 1
Reputation: 140287
map
applies the function in first parameter to the second parameter. It tries to call your filter
object...
A fixed version would be:
list(map(lambda z: list(filter(None,z)),lst))
(still not very clear even if it yields the expected result... it took me 3 or 4 tries to get it right by trial & error, and the fact that filter
& map
need to be iterated upon in Python 3 doesn't help and kills the whole hype about those fashionable keywords)
At this point of lambda/map/filter
complexity you'd be better off with a list comprehension:
lst = [ [33,3,3,0] , [34,1,0,4] ]
result = [ [x for x in sl if x] for sl in lst]
print(result)
result:
[[33, 3, 3], [34, 1, 4]]
Upvotes: 9
Reputation: 26331
The thing is that map
expects a function/callable as a first argument but you're giving it the result of a filter
.
You can fix this by applying filter
to the sublists iterated by map
:
def no_zeros(lst):
a = map(lambda x: filter(lambda y: y != 0, x), lst)
print(list(a))
By the way, you can do the same thing with list comprehensions:
def no_zeros(lst):
a = [[y for y in x if y != 0] for x in lst]
print(a)
Upvotes: 2
Reputation: 53089
Here is one possibility:
import itertools
a = [ [33,3,3,0] , [34,1,0,4] ]
list(map(list, map(filter, itertools.repeat(bool), a)))
# [[33, 3, 3], [34, 1, 4]]
Upvotes: 1
Reputation: 59
lst = list(map(lambda k: list(filter(lambda l: l != 0, k)), lst))
This should do the trick. You need apply a lambda on each element in the list through a filter. Be aware nested lambdas are ugly.
Upvotes: 1