Remove all matching value from a list of lists

Question

I have following list of lists object

myList = [[123,0.0,345,0.0,0.0,0.0],
 [45,0.0,0.0,0.0],
 [67,8,0.0,5,6,7,0.0]

And I want to remove all zeroes from this list.

I followed this question and coded like following.

myList = list(filter(lambda j:j!=0,myList[i]) for i in range(len(myList)))

But I am getting the list of filter object as the output. What is the error in the code.

[<filter object at 0x7fe7bdfff8d0>, <filter object at 0x7fe7a6eaaf98>, <filter object at 0x7fe7a6f08048>,

Answer 1

You forgot to cast the inner filter function with a list, when you do that, the code works as expected :)

myList = [[123,0.0,345,0.0,0.0,0.0],
 [45,0.0,0.0,0.0],
 [67,8,0.0,5,6,7,0.0]]

#Cast inner filter into a list
myList = list(list(filter(lambda j:j!=0,myList[i])) for i in range(len(myList)))
print(myList)

The output will be

[[123, 345], [45], [67, 8, 5, 6, 7]]

Also a simpler way of understanding will be to use a list-comprehension

myList = [[123,0.0,345,0.0,0.0,0.0],
 [45,0.0,0.0,0.0],
 [67,8,0.0,5,6,7,0.0]]

#Using list comprehension, in the inner loop check if item is non-zero
myList = [ [item for item in li if item != 0] for li in myList ]
print(myList)

The output will be

[[123, 345], [45], [67, 8, 5, 6, 7]]

Answer 2

Try this :

newList = [list(filter(lambda j:j!=0, i)) for i in myList]

OUTPUT :

[[123, 345], [45], [67, 8, 5, 6, 7]]

Answer 3

You could also do this with a list comprehension:

cleaned = [ [e for e in row if e != 0] for row in myList ]

Answer 4

You just need to wrap filter, not the whole statement:

myList = [list(filter(lambda j:j!=0,myList[i]) for i in range(len(myList))]

Also, you can skip the index, and iterate by lists in myList:

myList = [list(filter(lambda j:j!=0, inner_list) for inner_list in myList]