CODEWITHSUNDEEP
pythonlist
TheStrangeQuark
TheStrangeQuark

Reputation: 2405

replace all 0's in a list with numpy.nan

I have a list with 0's in it that I want to change to np.nan for calculations in doing. I am trying to do the following which only replaces the first 0:

l = [0,2,3,4,0]
l[l == 0] = np.nan
print l
[nan,2,3,4,0]

Is there a way to do this substitution for every item in the list?

Upvotes: 2

Views: 7593

Answers (4)

Nras
Nras

Reputation: 4311

Since you are already using numpy: your notation works, if l was a numpy.array and not a list. However, it only works, if your entries are not integers, e.g. floats:

array = np.array(list, dtype=float)
array[array == 0] = np.nan

Now array looks like this:

array([ nan, 2.,  3., 4., nan])

If you really need a list, then

l = list(array)

Upvotes: 1

JoeCondron
JoeCondron

Reputation: 8906

Here's a nice elegant way to do it:

l = [x or np.nan for x in l]

Upvotes: 0

DSM
DSM

Reputation: 352979

You can't use numpy vector boolean indexing or comparisons with a Python list, so your l == 0 is simply giving you 0. It's only a fluke it works even partially, because you happened to have 0 as the first element.

>>> l = [0,2,3,4,0]
>>> l == 0
False
>>> int(l == 0)
0
>>> l[l == 0] = 999
>>> l
[999, 2, 3, 4, 0]

You could use a list comprehension and a ternary:

>>> l = [0,2,3,4,0]
>>> l = [np.nan if x == 0 else x for x in l]
>>> l
[nan, 2, 3, 4, nan]

(If you were working with numpy arrays, vector comparisons and assignments would work, but you can't mix integers and floats (such as np.nan) too nicely.)

Upvotes: 7

bchoi
bchoi

Reputation: 126

You can just use the built in map:

l = map(lambda number: np.nan if number == 0 else number, l)

Which is equivalent to:

for i in range(len(l)):
    l[i] = 0 if l[i] == np.nan else l[i]

Upvotes: 5

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