Numpy equivalent of if/else without loop

Question

is there any pythonic way to remove for loop and if/else in the code below.

this code iterating over a NumPy array and check a condition and according to the condition change the value.

>>> import numpy as np
>>> x=np.random.randint(100, size=(10,5))
>>> x
array([[79, 50, 18, 55, 35],
       [46, 71, 46, 95, 52],
       [97, 37, 71,  2, 79],
       [80, 96, 60, 85, 72],
       [ 6, 52, 63, 86, 38],
       [35, 50, 13, 93, 54],
       [69, 21,  4, 40, 53],
       [83,  7, 30, 16, 78],
       [18, 34, 91, 67, 89],
       [82, 16, 16, 24, 80]])

>>> for i in range(x.shape[0]):
    for j in range(x.shape[1]):
        if x[i,j]>50:
            x[i,j]=0
        elif x[i,j]<50:
            x[i,j]=1


>>> x
array([[ 0, 50,  1,  0,  1],
       [ 1,  0,  1,  0,  0],
       [ 0,  1,  0,  1,  0],
       [ 0,  0,  0,  0,  0],
       [ 1,  0,  0,  0,  1],
       [ 1, 50,  1,  0,  0],
       [ 0,  1,  1,  1,  0],
       [ 0,  1,  1,  1,  0],
       [ 1,  1,  0,  0,  0],
       [ 0,  1,  1,  1,  0]])

I want to do same thing without loops and if statement.
something like below dose not work, because of changes on the array:

>>> import numpy as np
>>> x=np.random.randint(100, size=(10,5))
>>> x
array([[ 2, 88, 27, 67, 29],
       [62, 44, 62, 87, 32],
       [80, 95, 31, 30, 33],
       [14, 41, 40, 95, 27],
       [53, 30, 35, 22, 98],
       [90, 39, 74, 28, 73],
       [10, 71,  0, 11, 37],
       [28, 25, 83, 24, 93],
       [30, 70, 15,  5, 79],
       [69, 43, 85, 68, 53]])
>>> x[x>50]=0
>>> x[x<50]=1
>>> x
array([[1, 1, 1, 1, 1],
       [1, 1, 1, 1, 1],
       [1, 1, 1, 1, 1],
       [1, 1, 1, 1, 1],
       [1, 1, 1, 1, 1],
       [1, 1, 1, 1, 1],
       [1, 1, 1, 1, 1],
       [1, 1, 1, 1, 1],
       [1, 1, 1, 1, 1],
       [1, 1, 1, 1, 1]])

UPDATE
and what happend if there are more conditions like :

   >>> import numpy as np
    >>> x=np.random.randint(100, size=(10,5))
    >>> x
    array([[87, 99, 70, 32, 28],
           [38, 76, 89, 17, 34],
           [28,  1, 40, 34, 67],
           [45, 47, 69, 78, 89],
           [14, 81, 46, 71, 97],
           [39, 45, 36, 36, 25],
           [87, 28,  1, 46, 99],
           [27, 98, 37, 36, 84],
           [55,  2, 23, 29,  9],
           [34, 79, 49, 76, 48]])
    >>> for i in range(x.shape[0]):
        for j in range(x.shape[1]):
            if x[i,j]>90:
                x[i,j]=9
            elif x[i,j]>80:
                x[i,j]=8
            elif x[i,j]>70:
                x[i,j]=7
            elif x[i,j]>60:
                x[i,j]=6
            elif x[i,j]>50:
                x[i,j]=5
            elif x[i,j]>40:
                x[i,j]=4
            else:
                x[i,j]=0


    >>> x
    array([[8, 9, 6, 0, 0],
           [0, 7, 8, 0, 0],
           [0, 0, 0, 0, 6],
           [4, 4, 6, 7, 8],
           [0, 8, 4, 7, 9],
           [0, 4, 0, 0, 0],
           [8, 0, 0, 4, 9],
           [0, 9, 0, 0, 8],
           [5, 0, 0, 0, 0],
           [0, 7, 4, 7, 4]])

Asked By: pd shah

||

Source

Answer 1

One IF-ELIF

Approach #1 One approach –

keep_mask = x==50
out = np.where(x>50,0,1)
out[keep_mask] = 50

Approach #2 Alternatively, for in-situ edit –

replace_mask = x!=50
x[replace_mask] = np.where(x>50,0,1)[replace_mask]
# Or (x<=50).astype(int) in place of np.where(x>50,0,1)

Code-golf? If you actually want to play code-golf/one-liner –

(x<=50)+(x==50)*49

Multiple IF-ELIFs

Approach #1

For a bit more generic case involving more if-elif parts, we could make use of np.searchsorted –

out_x = np.where(x<=40,0, np.searchsorted([40,50,60,70,80,90], x)+3)

Answered By: Divakar

Answer 2

A one-liner that does everything your loops does:

x[x != 50] = x[x != 50] < 50

EDIT:

For your extended question, you’d want something like:

bins = [40, 50, 60, 70, 80, 90, 100]
out = np.digitize(x, bins, right = 1)
out[out.astype(bool)] += 3

Answered By: Daniel F

Answer 3

np.where(x < 50, 0, 1)

This should be enough. You don’t need to keep a mask value for 50 since 50 is neither less than nor greater than 50. Hope this helps.

Answered By: benjaminudoh10

Answer 4

np.where(x < 50, 0, 1)

This should be enough. You don’t need to keep a mask value for 50 since 50 is neither less than nor greater than 50. Hope this helps.

Update

#update
np.where(x < 40, 0, x)
np.where(x > (x - (x % 10)), x // 10, x)

Answered By: benjaminudoh10

Answer 5

Sorry for being late to the party, just wanted to share another approach to the problem.

One-Line Solution:

x = np.where(x>=50, 50, 1) + np.where(x>50, -50, 0)

Rationale:

We can sum over the following two numpy.where-matrices:

For matrix A: if x[i,j] >= 50, then set value 50, otherwise 1 because we want x[i,j]<50 to be equal to 1.
For matrix B: if x[i,j] > 50, then set value -50, thus for x[i,j]>50 the sum over both matrices will yield value 0 for the corresponding elements.

By calculating A+B, the values set for conditions x>50 (i.e. -50) and x>=50 (i.e. 50) yield the wanted values (0 and 50) and do not interfere with values set for x<50.

Update for UPDATE

x = np.where(x>40, 4, 0) + np.where(x>50, 1, 0) + np.where(x>60, 1, 0) + np.where(x>70, 1, 0) + np.where(x>80, 1, 0) + np.where(x>90, 1, 0)

Or shorter, if we can rely on the fact that values are always smaller than 100 (change dtype if you want integers):

x = np.where(x>40, np.floor(x/10), 0)

For me this code is quite readable, but I may not be representative.

Answered By: MarcelF

Numpy equivalent of if/else without loop

Question:

Answers:

One IF-ELIF

Multiple IF-ELIFs