CODEWITHSUNDEEP
pythonnumpylambdafunctional-programming
kakk11
kakk11

Reputation: 918

Use numpy array as lambda argument?

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]))

Upvotes: 2

Views: 2678

Answers (3)

user11553043
user11553043

Reputation:

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
--
--
--

Upvotes: 0

cabreracanal
cabreracanal

Reputation: 934

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 -- -- --]
>>>

Upvotes: 0

chepner
chepner

Reputation: 531245

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]))

Upvotes: 2

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