How to sort and array by indices sequence of another array

Question

Given I have two arrays:

array1 = np.array([np.nan,np.nan,np.nan,np.nan,2,np.nan,1,np.nan,np.nan,5,np.nan,np.nan,6,np.nan,10,9,np.nan])
array2 = np.array([np.nan,np.nan,np.nan,np.nan,45,np.nan,33,np.nan,np.nan,32,np.nan,np.nan,44,np.nan,10,53,np.nan])

I want to get array2 sorted in ascending order of array1 elements with same nan sequence: [np.nan,np.nan,np.nan,np.nan,32,np.nan,10,np.nan,np.nan,33,np.nan,np.nan,44,np.nan,53,45,np.nan]

Seems I could use np.argsort(array1) = [ 6 4 9 12 15 14 0 13 11 10 8 5 3 2 1 7 16] if there's a command to move elements in array2 like : put an element with index "6" to the first not nan place etc

Any ideas?

UPD1: I posted a related question How to replace elements of a numpy array from two different arrays for a case when we have to split the array for reordering

Akshay Sehgal · Accepted Answer · 2021-10-12 13:45:24Z

2

You can do this with numpy by overwriting the view over the second array for non nan elements.

idx = array1[~np.isnan(array1)].argsort()

array2[~np.isnan(array2)] = array2[~np.isnan(array2)][idx]
array2

array([nan, nan, nan, nan, 33., nan, 45., nan, nan, 32., nan, nan, 44.,
       nan, 53., 10., nan])

Another "cleaner" way to do this is to just sort the arrays as is, remove nans and then overwrite the view for array2 as in the previous approach.

ordered = array2[array1.argsort()]
ordered = ordered[~np.isnan(ordered)]

array2[~np.isnan(array2)] = ordered
array2

array([nan, nan, nan, nan, 33., nan, 45., nan, nan, 32., nan, nan, 44.,
       nan, 53., 10., nan])

edited Oct 12, 2021 at 13:45

answered Oct 12, 2021 at 13:25

Akshay Sehgal

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8 Comments

Nikolay Yasinskiy Over a year ago

I loved the cleaner way, thanks!

Akshay Sehgal Over a year ago

glad to help anytime.

Stef Over a year ago

For the second way, you sort first, then remove the NaNs. Wouldn't it be more efficient to remove the NaNs first, then sort?

Akshay Sehgal Over a year ago

correct, thats why its just the "cleaner" way not the more efficient one (which the first one is) :)

Nikolay Yasinskiy Over a year ago

Euh, MemoryError: Unable to allocate array with shape (5926, 6331, 6331) and data type float32 for ordered = array2[array1.argsort()]. Isn'tappropriate for my big fields unfortunately.

|

Stef · Accepted Answer · 2021-10-12 13:42:45Z

2

Remove NaN, sort, add NaN.

from operator import itemgetter

#remove NaN
a = [x for x in array1 if not np.isnan(x)]
b = [y for y in array2 if not np.isnan(y)]
# a == [2.0, 1.0, 5.0, 6.0, 10.0, 9.0]
# b == [45.0, 33.0, 32.0, 44.0, 10.0, 53.0]


# sort
c = map(itemgetter(1), sorted(zip(a, b)))
# list(c) == [33.0, 45.0, 32.0, 44.0, 53.0, 10.0]

# add NaN
d = [(np.nan if np.isnan(x) else next(c)) for x in array1]
# d == [nan, nan, nan, nan, 33.0, nan, 45.0, nan, nan, 32.0, nan, nan, 44.0, nan, 53.0, 10.0, nan]

Note that the suggestion in Akshay Sehgal's answer is likely faster than mine, since it uses numpy directly rather than python lists and iterators.

edited Oct 12, 2021 at 13:42

answered Oct 12, 2021 at 13:22

Stef

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3 Comments

Akshay Sehgal Over a year ago

interesting way of using itermgetter

Stef Over a year ago

@AkshaySehgal Instead I could write c = (y for _,y in sorted(zip(a,b))), but I like map :). The important point is that c is an iterator, not a list.

Akshay Sehgal Over a year ago

python is as much about readability as it is about computation time, so either work :)

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