I have a list of strings
x = ['A', 'B', nan, 'D']
and want to remove the nan.
I tried:
x = x[~numpy.isnan(x)]
But that only works if it contains numbers. How do we solve this for strings in Python 3+?
4 Answers
If you have a numpy array you can simply check the item is not the string nan, but if you have a list you can check the identity with is and np.nan since it's a singleton object.
In [25]: x = np.array(['A', 'B', np.nan, 'D'])
In [26]: x
Out[26]:
array(['A', 'B', 'nan', 'D'],
dtype='<U3')
In [27]: x[x != 'nan']
Out[27]:
array(['A', 'B', 'D'],
dtype='<U3')
In [28]: x = ['A', 'B', np.nan, 'D']
In [30]: [i for i in x if i is not np.nan]
Out[30]: ['A', 'B', 'D']
Or as a functional approach in case you have a python list:
In [34]: from operator import is_not
In [35]: from functools import partial
In [37]: f = partial(is_not, np.nan)
In [38]: x = ['A', 'B', np.nan, 'D']
In [39]: list(filter(f, x))
Out[39]: ['A', 'B', 'D']
6 Comments
Colonel Beauvel
aggregate things like:
[i for i in x if not i in ['nan', np.nan]], +1 otherwise
Kasravnd
@ColonelBeauvel Yeah, that's a good idea if you don't know that kind of a data structure you're dealing with.
Josh Lee
NaN is not a singleton.
Kasravnd
@JoshLee Why? I think as far as you can't create different instances from a particular object it would be refer as singleton. Is there anything special about
np.nan?Josh Lee
np.nan is just some floating point constant. You wouldn't compare is math.pi either, for the same reason. |
You can use math.isnan and a good-old list comprehension.
Something like this would do the trick:
import math
x = [y for y in x if not math.isnan(y)]
1 Comment
hpaulj
Did you try
math.isnan('A')? Test on the OP's x?You may want to avoid np.nan with strings, use None instead; but if you do have nan you could do this:
import numpy as np
[i for i in x if i is not np.nan]
# ['A', 'B', 'D']
4 Comments
Josh Lee
NaN is not a singleton.
akuiper
@JoshLee I didn't say it's a singleton. I just said it might be better to use None instead of
nan in string cases, which will be converted to a string nan but None stays as None.Josh Lee
You're comparing with
is. This will fail.akuiper
@JoshLee I didn't get what you mean. It works for this case.
You could also try this:
[s for s in x if str(s) != 'nan']
Or, convert everything to str at the beginning:
[s for s in map(str, x) if s != 'nan']
Both approaches yield ['A', 'B', 'D'].
nonobject from numpy module which the OP is using. I change it to numpy so that the future askers can find the question easily.