Use numpy array as lambda argument?

Question

Is there a reasonable way to get the following done on one line? I'd really like to avoid creating a temporary variable or a separate function.

import numpy as np
x = np.array([1,2,3,4,5])
x = np.ma.masked_where(x>2, x)

I tried

x = map(lambda x: np.ma.masked_where(x>2, x), np.array([1,2,3,4,5]))

but the map object is not what I want? I can of course define separate fuction, which avoids assigning variable:

masker = lambda x: np.ma.masked_where(x>2, x)
x = masker(np.array([1,2,3,4,5]))

In your second "attempt", you would still be creating the intermediate array, you just aren't assigning it. I'm confused why the first approach isn't what you want — user3483203
– user3483203, Commented Jul 24, 2019 at 17:02
I understand I cannot avoid creating an intermediate array, just want to avoid assigning it. Why, just trying to look for ways to be more functional with numpy. — kakk11
– kakk11, Commented Jul 24, 2019 at 17:10
The assignment of the intermediate array causes absolutely no problems, because you immediately reassign the same variable with the final result. — jasonharper
– jasonharper, Commented Jul 24, 2019 at 17:16

chepner · Accepted Answer · 2019-07-24 17:18:33Z

2

You don't need map at all, just an anonymous function. All you will do is replace the initial assignment to x with a parameter binding in a function call.

import numpy as np
# x = np.array([1,2,3,4,5])
# x = np.ma.masked_where(x>2, x)

x = (lambda x: np.ma.masked_where(x>2, x))(np.array([1,2,3,4,5]))

answered Jul 24, 2019 at 17:18

chepner

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1 Comment

kakk11 Over a year ago

Thanks, I guess that's what I was looking for, but couldn't figure out the syntax.

cabreracanal · Accepted Answer · 2019-07-24 17:22:39Z

0

This works great for me and is one liner:

>>> x = (lambda y: np.ma.masked_where(y>2, y))(np.array([1,2,3,4,5]))
>>> print (x)
[1 2 -- -- --]
>>>

answered Jul 24, 2019 at 17:22

cabreracanal

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2 Comments

rafaelc Over a year ago

How is this different from chepner's answer? :)

cabreracanal Over a year ago

It is not. I didn't refresh the page while testing on my computer and posting the answer.

score 0 · Accepted Answer · 2019-07-24 17:40:51Z

0

Here is a way to do this with map:

import numpy as np

x = map(
    np.ma.masked_where, 
    *(np.array([1,2,3,4,5])>2, np.array([1,2,3,4,5]))
)

Map returns an iterable, so to review the masking, go like:

>>> for item in x:
...     print(item)
... 
1
2
--
--
--

edited Jul 24, 2019 at 17:40

answered Jul 24, 2019 at 17:32

user11553043

1 Comment

kakk11 Over a year ago

Interesting, but I guess iteration over individual elements is not what one usually does with numpy arrays.

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