how to apply a mask from one array to another array?

Question

I've read the masked array documentation several times now, searched everywhere and feel thoroughly stupid. I can't figure out for the life in me how to apply a mask from one array to another.

Example:

import numpy as np

y = np.array([2,1,5,2])          # y axis
x = np.array([1,2,3,4])          # x axis
m = np.ma.masked_where(y>2, y)   # filter out values larger than 5
print m
[2 1 -- 2]
print np.ma.compressed(m)
[2 1 2]

So this works fine.... but to plot this y axis, I need a matching x axis. How do I apply the mask from the y array to the x array? Something like this would make sense, but produces rubbish:

new_x = x[m.mask].copy()
new_x
array([5])

So, how on earth is that done (note the new x array needs to be a new array).

Edit:

Well, it seems one way to do this works like this:

>>> import numpy as np
>>> x = np.array([1,2,3,4])
>>> y = np.array([2,1,5,2])
>>> m = np.ma.masked_where(y>2, y)
>>> new_x = np.ma.masked_array(x, m.mask)
>>> print np.ma.compressed(new_x)
[1 2 4]

But that's incredibly messy! I'm trying to find a solution as elegant as IDL...

Can't you just plot like plot(x, m) without making a new_x? — joris
– joris, Commented May 11, 2013 at 8:40
And it is new_x = x[~m.mask].copy(). Note the ~, as the mask is True where the value is masked. — joris
– joris, Commented May 11, 2013 at 8:41
Note that there is no need for .copy() in new_x = x[~m.mask].copy(). Indexing with a boolean array will always result in a copy, so this can be new_x = x[~m.mask]. — Warren Weckesser
– Warren Weckesser, Commented May 11, 2013 at 9:54
Do you really need to use mask here? Indexing over boolean array will give you what you want. new_y = y[y < 5] and new_x = x[y < 5]. — Akavall
– Akavall, Commented May 11, 2013 at 12:29
You can just assign that to a variable if you want to reuse it: mask = y < 5; m = y[mask]; new_x = x[mask]. In this way you avoid using masked arrays and keep it more simple and straight. — joris
– joris, Commented May 11, 2013 at 15:29

red_tiger · Accepted Answer · 2013-08-15 12:18:54Z

28

I had a similar issue, but involving loads more masking commands and more arrays to apply them. My solution is that I do all the masking on one array and then use the finally masked array as the condition in the mask_where command.

For example:

y = np.array([2,1,5,2])                         # y axis
x = np.array([1,2,3,4])                         # x axis
m = np.ma.masked_where(y>5, y)                  # filter out values larger than 5
new_x = np.ma.masked_where(np.ma.getmask(m), x) # applies the mask of m on x

The nice thing is you can now apply this mask to many more arrays without going through the masking process for each of them.

edited Aug 15, 2013 at 12:18

answered Aug 15, 2013 at 9:53

red_tiger

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Comments

kiriloff · Accepted Answer · 2013-05-11 09:06:57Z

22

Why not simply

import numpy as np

y = np.array([2,1,5,2])          # y axis
x = np.array([1,2,3,4])          # x axis
m = np.ma.masked_where(y>2, y)   # filter out values larger than 5
print list(m)
print np.ma.compressed(m)

# mask x the same way
m_ = np.ma.masked_where(y>2, x)   # filter out values larger than 5
# print here the list
print list(m_) 
print np.ma.compressed(m_)

code is for Python 2.x

Also, as proposed by joris, this do the work new_x = x[~m.mask].copy() giving an array

>>> new_x
array([1, 2, 4])

answered May 11, 2013 at 9:06

kiriloff

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3 Comments

Balthasar Over a year ago

Good stuff also, thanks. I'm obviously new to Python and suffering my way through...

Marshall Moutenot Over a year ago

It may help your answers to avoid the words "simply," "just" etc.

Eduardo Pignatelli Over a year ago

What if x and y have different number of dimensions and I want to mask only on axis 0?

syoels · Accepted Answer · 2020-11-23 18:30:05Z

This may not bee 100% what OP wanted to know, but it's a cute little piece of code I use all the time - if you want to mask several arrays the same way, you can use this generalized function to mask a dynamic number of numpy arrays at once:

def apply_mask_to_all(mask, *arrays):
assert all([arr.shape == mask.shape for arr in arrays]), "All Arrays need to have the same shape as the mask"
return tuple([arr[mask] for arr in arrays])

See this example usage:

    # init 4 equally shaped arrays
x1 = np.random.rand(3,4)
x2 = np.random.rand(3,4)
x3 = np.random.rand(3,4)
x4 = np.random.rand(3,4)

# create a mask
mask = x1 > 0.8

# apply the mask to all arrays at once
x1, x2, x3, x4 = apply_mask_to_all(m, x1, x2, x3, x4)

learner · Accepted Answer · 2024-10-19 19:05:42Z

0

I know it is an old post, but for those who may want to accomplish something similar in a probably more straightforward way:

Approach 1:

import numpy as np

y = np.array([2,1,5,2])
x = np.array([1,2,3,4])

locs=~(y>2)
new_x=x[locs]
new_y=y[locs]

print('locs: ', locs)
print('new_x: ', new_x)
print('new_y: ', new_y)

Output:

locs:  [ True  True False  True]
new_x:  [1 2 4]
new_y:  [2 1 2]

Approach 2:

import numpy as np

y = np.array([2,1,5,2])
x = np.array([1,2,3,4])

locs=np.where(~(y>2))[0]
new_x=x[locs]
new_y=y[locs]

print('locs: ', locs)
print('new_x: ', new_x)
print('new_y: ', new_y)

Output:

locs:  [0 1 3]
new_x:  [1 2 4]
new_y:  [2 1 2]

answered Oct 19, 2024 at 19:05

learner

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