Question

我有一个非常大的numpy数组（包含多达一百万个元素），如下所示：

[ 0  1  6  5  1  2  7  6  2  3  8  7  3  4  9  8  5  6 11 10  6  7 12 11  7
  8 13 12  8  9 14 13 10 11 16 15 11 12 17 16 12 13 18 17 13 14 19 18 15 16
 21 20 16 17 22 21 17 18 23 22 18 19 24 23]

和一个小字典图，用于替换上面数组中的一些元素

{4: 0, 9: 5, 14: 10, 19: 15, 20: 0, 21: 1, 22: 2, 23: 3, 24: 0}

我想根据上面的地图替换一些元素。 numpy数组非常大，只有一小部分元素（作为字典中的键出现）将被相应的值替换。最快的方法是什么？

Answer 1

我相信有更有效的方法，但是现在，试试

from numpy import copy

newArray = copy(theArray)
for k, v in d.iteritems(): newArray[theArray==k] = v

Microbenchmark并测试其正确性：

#!/usr/bin/env python2.7

from numpy import copy, random, arange

random.seed(0)
data = random.randint(30, size=10**5)

d = {4: 0, 9: 5, 14: 10, 19: 15, 20: 0, 21: 1, 22: 2, 23: 3, 24: 0}
dk = d.keys()
dv = d.values()

def f1(a, d):
    b = copy(a)
    for k, v in d.iteritems():
        b[a==k] = v
    return b

def f2(a, d):
    for i in xrange(len(a)):
        a[i] = d.get(a[i], a[i])
    return a

def f3(a, dk, dv):
    mp = arange(0, max(a)+1)
    mp[dk] = dv
    return mp[a]


a = copy(data)
res = f2(a, d)

assert (f1(data, d) == res).all()
assert (f3(data, dk, dv) == res).all()

结果：

$ python2.7 -m timeit -s 'from w import f1,f3,data,d,dk,dv' 'f1(data,d)'
100 loops, best of 3: 6.15 msec per loop

$ python2.7 -m timeit -s 'from w import f1,f3,data,d,dk,dv' 'f3(data,dk,dv)'
100 loops, best of 3: 19.6 msec per loop

Answer 2

假设值介于0和某个最大整数之间，可以使用numpy-array作为int->int dict实现快速替换，如下所示

mp = numpy.arange(0,max(data)+1)
mp[replace.keys()] = replace.values()
data = mp[data]

首先

data = [ 0  1  6  5  1  2  7  6  2  3  8  7  3  4  9  8  5  6 11 10  6  7 12 11  7
  8 13 12  8  9 14 13 10 11 16 15 11 12 17 16 12 13 18 17 13 14 19 18 15 16
 21 20 16 17 22 21 17 18 23 22 18 19 24 23]

并替换为

replace = {4: 0, 9: 5, 14: 10, 19: 15, 20: 0, 21: 1, 22: 2, 23: 3, 24: 0}

我们获得

data = [ 0  1  6  5  1  2  7  6  2  3  8  7  3  0  5  8  5  6 11 10  6  7 12 11  7
  8 13 12  8  5 10 13 10 11 16 15 11 12 17 16 12 13 18 17 13 10 15 18 15 16
  1  0 16 17  2  1 17 18  3  2 18 15  0  3]

Answer 3

另一种更通用的方法是函数矢量化：

import numpy as np

data = np.array([0, 1, 6, 5, 1, 2, 7, 6, 2, 3, 8, 7, 3, 4, 9, 8, 5, 6, 11, 10, 6, 7, 12, 11, 7, 8, 13, 12, 8, 9, 14, 13, 10, 11, 16, 15, 11, 12, 17, 16, 12, 13, 18, 17, 13, 14, 19, 18, 15, 16, 21, 20, 16, 17, 22, 21, 17, 18, 23, 22, 18, 19, 24, 23])
mapper_dict = {4: 0, 9: 5, 14: 10, 19: 15, 20: 0, 21: 1, 22: 2, 23: 3, 24: 0}

def mp(entry):
    return mapper_dict[entry] if entry in mapper_dict else entry
mp = np.vectorize(mp)

print mp(data)

Answer 4

没有在阵列上没有python循环的情况下仍未发布任何解决方案（除了Celil＆＃39;然而，假设数字为＆＃34;小＆＃34;），所以这里有另一种选择：

def replace(arr, rep_dict):
    """Assumes all elements of "arr" are keys of rep_dict"""

    # Removing the explicit "list" breaks python3
    rep_keys, rep_vals = array(list(zip(*sorted(rep_dict.items()))))

    idces = digitize(arr, rep_keys, right=True)
    # Notice rep_keys[digitize(arr, rep_keys, right=True)] == arr

    return rep_vals[idces]

方式＆＃34; idces＆＃34;创建来自here。

Answer 5

我对一些解决方案进行了基准测试，结果没有吸引力：

import timeit
import numpy as np

array = 2 * np.round(np.random.uniform(0,10000,300000)).astype(int)
from_values = np.unique(array) # pair values from 0 to 2000
to_values = np.arange(from_values.size) # all values from 0 to 1000
d = dict(zip(from_values, to_values))

def method_for_loop():
    out = array.copy()
    for from_value, to_value in zip(from_values, to_values) :
        out[out == from_value] = to_value
    print('Check method_for_loop :', np.all(out == array/2)) # Just checking
print('Time method_for_loop :', timeit.timeit(method_for_loop, number = 1))

def method_list_comprehension():
    out = [d[i] for i in array]
    print('Check method_list_comprehension :', np.all(out == array/2)) # Just checking
print('Time method_list_comprehension :', timeit.timeit(method_list_comprehension, number = 1))

def method_bruteforce():
    idx = np.nonzero(from_values == array[:,None])[1]
    out = to_values[idx]
    print('Check method_bruteforce :', np.all(out == array/2)) # Just checking
print('Time method_bruteforce :', timeit.timeit(method_bruteforce, number = 1))

def method_searchsort():
    sort_idx = np.argsort(from_values)
    idx = np.searchsorted(from_values,array,sorter = sort_idx)
    out = to_values[sort_idx][idx]
    print('Check method_searchsort :', np.all(out == array/2)) # Just checking
print('Time method_searchsort :', timeit.timeit(method_searchsort, number = 1))

我得到了以下结果：

Check method_for_loop : True
Time method_for_loop : 2.6411612760275602

Check method_list_comprehension : True
Time method_list_comprehension : 0.07994363596662879

Check method_bruteforce : True
Time method_bruteforce : 11.960559037979692

Check method_searchsort : True
Time method_searchsort : 0.03770717792212963

“searchsort”方法比“for”循环快一百次，并且比numpy bruteforce方法快3600倍。列表理解方法也是代码简单性和速度之间的非常好的折衷。

Answer 6

numpy_indexed包（免责声明：我是它的作者）为这类问题提供了一种优雅而高效的矢量化解决方案：

import numpy_indexed as npi
remapped_array = npi.remap(theArray, list(dict.keys()), list(dict.values()))

实施的方法类似于Jean Lescut提到的基于搜索排序的方法，但更为一般。例如，数组的项不需要是int，但可以是任何类型，甚至是nd-subarrays本身;但它应该达到同样的性能。

Answer 7

好吧，你需要通过theArray进行一次传递，并且如果它在字典中，则替换每个元素。

for i in xrange( len( theArray ) ):
    if foo[ i ] in dict:
        foo[ i ] = dict[ foo[ i ] ]

Answer 8

for i in xrange(len(the_array)):
    the_array[i] = the_dict.get(the_array[i], the_array[i])

Answer 9

Pythonic方式无需数据为整数，甚至可以是字符串：

from scipy.stats import rankdata
import numpy as np

data = np.random.rand(100000)
replace = {data[0]: 1, data[5]: 8, data[8]: 10}

arr = np.vstack((replace.keys(), replace.values())).transpose()
arr = arr[arr[:,1].argsort()]

unique = np.unique(data)
mp = np.vstack((unique, unique)).transpose()
mp[np.in1d(mp[:,0], arr),1] = arr[:,1]
data = mp[rankdata(data, 'dense')-1][:,1]

快速替换numpy数组中的值

9 个答案: