Filtering Lists in a List of Lists (Python) Based on Values

Question

Hey I'm trying to write a function that only prints a list that contains no zeros in a list of lists. Here is a sample of the list of lists I am working with:

[['10011.0', ' 65301.0', ' 0.085', ' 0.0', ' 0.0', ' 0.0', ' 0.03', ' 0.03', ' 0.075',
 ' 0.05', ' 0.065', ' 0.05', ' 0.05', ' 0.0', ' 0.0', ' 0.0', ' 0.0', ' 0.0', ' 0.0'],
 ['10017.0', ' 2743.0', ' 0.0', ' 0.0', ' 0.0', ' 0.0', ' 0.0', ' 0.0', ' 0.0', ' 0.0', ' 0.0', ' 0.0', ' 0.2', ' 0.413333', ' 0.415', ' 0.3125', ' 0.45', ' 0.46', ' 0.55'],
 ['10017.0', ' 9262.0', ' 0.0', ' 0.0', ' 0.0', ' 0.0', ' 0.0', ' 0.0', ' 0.0', ' 0.0', ' 0.0', ' 0.0', ' 0.35', ' 0.69', ' 0.675', ' 0.8075', ' 0.8075', ' 0.5325', ' 0.785'], 
 ['10017.0', ' 29319.0', ' 0.0', ' 0.0', ' 0.0', ' 0.0', ' 0.0', ' 0.0', ' 0.0', ' 0.0', ' 0.0', ' 0.0', ' 0.0225', ' 0.06', ' 0.0575', ' 0.105', ' 0.1', ' 0.045', ' 0.0'],
 ['10017.0', ' 43562.0', ' 0.0', ' 0.0', ' 0.0', ' 0.0', ' 0.0', ' 0.0', ' 0.0', ' 0.0', ' 0.0', ' 0.0', ' 0.0', ' 0.106667', ' 0.0925', ' 0.09', ' 0.1', ' 0.09', ' 0.1025'],
 ['10017.0', ' 43563.0', ' 1.0', ' 1.0', ' 1.0', ' 1.0', ' 1.0', ' 1.0', ' 1.0', ' 1.0', ' 1.0', ' 1.0', ' 1.0', ' 0.106667', ' 0.0925', ' 0.09', ' 0.1', ' 0.09', ' 0.1028']]

The code I am running is:

def no_zero(A):
for i in range(len(A)):
    for j in range(2, (len(A[i])+1)):
        if '0.0' not in A[i]:
            print A[i]
            break

For some reason it does not filter any of the lists and prints the entire list of lists despite the conditional: "if '0.0' not in A[i]:" I am unsure where my error is because it seems like the logic should be fairly straightforward. Thanks for any help!

Why do you have the second for loop? You never use j anywhere. — TheSoundDefense
– TheSoundDefense, Commented Jul 20, 2014 at 3:06
I need it to skip the first two elements of the list. Sorry for not clarifying that in the initial question. — user3758443
– user3758443, Commented Jul 20, 2014 at 3:10
Alright, but you still never use j anywhere in your code. You also don't want to use len+1 as your termination point, because A[len(A)] never exists. — TheSoundDefense
– TheSoundDefense, Commented Jul 20, 2014 at 3:12

thefourtheye · Accepted Answer · 2014-07-20 03:24:37Z

3

Normally when you want to filter a list of items, we use list comprehension with the filtering condition to get a new list. So, in your case, it can be done like this

[c_list for c_list in list_of_lists if '0.0' not in c_list]

This will give all the lists with no 0.0 in them. If you want to skip the first two elements, then start from the third element, while checking

[c_list for c_list in list_of_lists if '0.0' not in c_list[2:]]

edited Jul 20, 2014 at 3:24

answered Jul 20, 2014 at 3:10

thefourtheye

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2 Comments

TheSoundDefense Over a year ago

It should be [2:], not [3:]. Lists are zero-indexed.

msw Over a year ago

Except the zeros in his lists are ' 0.0' and I'd not count on that particular spacing. `Tis better to use floats; at least 0.0 is guaranteed precise representation in IEEE-754.

msw · Accepted Answer · 2014-07-20 05:23:29Z

1

You are making this much harder than it needs to be.

List indices are often unnecessary in Python; use iteration instead.
Testing strings for equality to '0.0' is troublesome when your data has spaces ' 0.0'
List comprehensions and built-ins are your friends.

Thus:

 def no_zeros(rows):
    """Takes a list of rows, prints each row if and only if it has no zeros."""
    for row in rows:
        # skip first two elements of row, per specification
        tests = [float(x) != 0 for x in row[2:]]
        if all(tests):
            print row

yields:

>>> no_zeros(A)
['10017.0', ' 43563.0', ' 1.0', ' 1.0', ' 1.0', ' 1.0', ' 1.0', ' 1.0', ' 1.0', 
 ' 1.0', ' 1.0', ' 1.0', ' 1.0', ' 0.106667', ' 0.0925', ' 0.09', ' 0.1', 
 ' 0.09', ' 0.1028']

edited Jul 20, 2014 at 5:23

answered Jul 20, 2014 at 5:12

msw

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Comments

NoobieGuy · Accepted Answer · 2014-07-20 08:55:35Z

1

Noobie attempt. This is if you want to test if the value is 0.

Whereby l is your list in list.

for i in l:
    a=[]
    for j in i[2:]:
        if float(j)==0:
          break
    else:
        a.append(i)

edited Jul 20, 2014 at 8:55

answered Jul 20, 2014 at 4:25

NoobieGuy

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2 Comments

msw Over a year ago

+1 pretty good for a noob. You even got the for / break / else thing right. You missed the for j in i[2:]: thing but that was hidden in a comment, not in the OP question.

msw Over a year ago

Nah, I missed that "skip the first two elements" myself at first. Be kind to yourself ;)

David Neiss · Accepted Answer · 2014-07-20 03:14:45Z

0

If A is your list of lists, filter(all, A)

answered Jul 20, 2014 at 3:14

David Neiss

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1 Comment

msw Over a year ago

Did you test it? A a-s an iterable is a list of length 6 (each of which is a list itself. A non-empty list is "trueish" so each of those lists in A will be true for all thus a list of all 6 lists will be returned.

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