如何根据另一个 NumPy 数组的值创建 NumPy 数组？

Question

I would like to create a NumPy array. The value of it's elements depends on the value of the elements in another NumPy array. Presently, I have to use a for-loop in a list comprehension to iterate through array a to get b . What is the NumPy way to achieve this?

Test Script:

import numpy as np

def get_b( a ):
    b_dict = {  1:10., 2:20., 3:30. }
    return b_dict[ a ]

a = np.full( 10, 2 )
print( f'a = {a}' )
b = np.array( [get_b(i) for i in a] )
print( f'b = {b}' )

Output:

a = [2 2 2 2 2 2 2 2 2 2]
b = [20. 20. 20. 20. 20. 20. 20. 20. 20. 20.]

Answer 1

You can use np.vectorize to map a dictionary value to an array

In [6]: b_dict = {  1:10., 2:20., 3:30 }

In [7]: a = np.full( 10, 2 )

In [8]: np.vectorize(b_dict.get)(a)
Out[8]: array([20., 20., 20., 20., 20., 20., 20., 20., 20., 20.])

Answer 2

What about using map and np.fromiter ?

def get_b( a ):
    b_dict = {  1:10., 2:20., 3:30. }
    return b_dict[ a ]

a = np.full( 10, 2 )
b = np.fromiter(map(get_b, a), dtype=np.float64)

Edit 1 : Small time comparison:

%timeit np.array( [get_b(i) for i in a] )
5.58 µs ± 123 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

%timeit np.fromiter(map(get_b, a), dtype=np.float64)
5.77 µs ± 177 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

%timeit np.vectorize(b_dict.get)(a)
12.9 µs ± 76.8 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

Edit 2 : Seems like that example is too small:

a = np.full( 1000, 2 )

%timeit np.array( [get_b(i) for i in a] )
415 µs ± 9.13 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

%timeit np.fromiter(map(get_b, a), dtype=np.float64)
383 µs ± 2.5 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

%timeit np.vectorize(b_dict.get)(a)
68.6 µs ± 625 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)

Answer 3

Another approach to the problem:

from operator import itemgetter
np.array(itemgetter(*a)(b_dict))

output:

[20., 20., 20., 20., 20., 20., 20., 20., 20., 20.]

---

Comparison :

#@kmundnic solution
def m1(a):
  def get_b(x):
    b_dict = {  1:10., 2:20., 3:30. }
    return b_dict[x]
  return np.fromiter(map(get_b, a),dtype=np.float)

#@bigbounty solution
def m2(a):
  b_dict = {  1:10., 2:20., 3:30. }
  return np.vectorize(b_dict.get)(a)

#@Ehsan solution
def m3(a):
  b_dict = {  1:10., 2:20., 3:30. }
  return np.array(itemgetter(*a)(b_dict))

#@Sun Bear solution
def m4(a):
  def get_b( a ):
    b_dict = {  1:10., 2:20., 3:30. }
    return b_dict[ a ]
  return np.array( [get_b(i) for i in a] )

in_ = [np.full( n, 2 ) for n in [10,100,1000,10000]]

For small dictionary , seems m2 is fastest at large inputs and m3 for smaller ones.

And for a larger dictionary :

b_dict = dict(zip(np.arange(100),np.arange(100)))
in_ = [np.full(n,50) for n in [10,100,1000,10000]]

m3 is the fastest approach. You can choose based on your dictionary size and key array size.

Answer 4

I like to stress the value of @hpaulj comment to my question:

Does b_dict have to be a dict? If you had an array, eg. ref = np.array([0, 10,20,30]) you quickly select the values by index, ref[a] . I would try to avoid dict when working with numpy.

I found that using NumPy's indexing will lead to a few to several orders of magnitude faster in performance than when trying to work with a python dict .

Building on @Ehsan's solution , below is a script that makes such a comparison.

import numpy as np
from operator import itemgetter
import timeit
import matplotlib.pyplot as plt


#@kmundnic solution
def m1(a):
    def get_b(x):
        b = {  1:10., 2:20., 3:30. }
        #b = dict( zip( np.arange(1,101),np.arange(10,1001,10) ) )
        return b[x]
    return np.fromiter(map(get_b, a),dtype=np.float)

#@bigbounty solution
def m2(a):
    b = {  1:10., 2:20., 3:30. }
    #b = dict( zip( np.arange(1,101),np.arange(10,1001,10) ) )
    return np.vectorize(b.get)(a)

#@Ehsan solution
def m3(a):
    b = {  1:10., 2:20., 3:30. }
    #b = dict( zip( np.arange(1,101),np.arange(10,1001,10) ) )
    return np.array(itemgetter(*a)(b))

#@Sun Bear solution
def m4(a):
    def get_b( a ):
        b = {  1:10., 2:20., 3:30. }
        #b = dict( zip( np.arange(1,101),np.arange(10,1001,10) ) )
        return b[ a ]
    return np.array( [get_b(i) for i in a] )

#@hpaulj solution
def m5(a):
    b = np.array([10, 20, 30])
    #b = np.arange(10,1001,10) 
    return b[a]

        
sizes=[10,100,1000,10000]
pm1 = []
pm2 = []
pm3 = []
pm4 = []
pm5 = []
for size in sizes:
    a = np.full( size, 2 )
    pm1.append( timeit.timeit( 'm1(a)', number=1000, globals=globals() ) )
    pm2.append( timeit.timeit( 'm2(a)', number=1000, globals=globals() ) )
    pm3.append( timeit.timeit( 'm3(a)', number=1000, globals=globals() ) )
    pm4.append( timeit.timeit( 'm4(a)', number=1000, globals=globals() ) )
    pm5.append( timeit.timeit( 'm5(a)', number=1000, globals=globals() ) )

print( 'm1 slower than m5 by :',np.array(pm1) / np.array(pm5) )
print( 'm2 slower than m5 by :',np.array(pm2) / np.array(pm5) )
print( 'm3 slower than m5 by :',np.array(pm3) / np.array(pm5) )
print( 'm4 slower than m5 by :',np.array(pm4) / np.array(pm5) )

fig = plt.figure()
ax = fig.add_subplot(1, 1, 1)
ax.plot( sizes, pm1, label='m1' )
ax.plot( sizes, pm2, label='m2' )
ax.plot( sizes, pm3, label='m3' )
ax.plot( sizes, pm4, label='m4' )
ax.plot( sizes, pm5, label='m5' )
ax.grid( which='both' )
ax.set_xscale('log')
ax.set_yscale('log')
ax.legend()
ax.get_xaxis().set_label_text( label='len(a)', fontweight='bold' )
ax.get_yaxis().set_label_text( label='Runtime (sec)', fontweight='bold' )
plt.show()

Results:

len(b) = 3:

m1 slower than m5 by : [  4.22462367  29.79407905  85.03454097 339.2915358 ]
m2 slower than m5 by : [  8.64220685 11.57175871 13.76761749 46.1940683 ]
m3 slower than m5 by : [  3.25785432  21.63131578  54.71305704 220.15777696 ]
m4 slower than m5 by : [  4.60710166  30.93616607  91.8936744  371.00398273 ]

len(b) = 100:

m1 slower than m5 by : [  218.98603678  1976.50128737  9697.76615006 17742.79151719 ]
m2 slower than m5 by : [  41.76535891  53.85600913 109.35129345 164.13075291 ]
m3 slower than m5 by : [  24.82715462  36.77830986  87.56253196 141.04493237 ]
m4 slower than m5 by : [  222.04184193  2001.72120836  9775.22464369 18431.00155305 ]