是否可以使用numpy乘法将字符串乘以整数？

Question

我尝试将两个数组按元素相乘，形成一个字符串。你知道吗

有人能建议吗？你知道吗

import numpy as np


def array_translate(array):

    intlist = [x for x in array if isinstance(x, int)]
    strlist = [x for x in array if isinstance(x, str)]
    joinedlist = np.multiply(intlist, strlist)
    return "".join(joinedlist)


print(array_translate(["Cat", 2, "Dog", 3, "Mouse", 1]))    # => "CatCatDogDogDogMouse"

我收到这个错误：

File "/Users/peteryoon/PycharmProjects/Test3/Test3.py", line 8, in array_translate
    joinedlist = np.multiply(intlist, strlist)
numpy.core._exceptions.UFuncTypeError: ufunc 'multiply' did not contain a loop with signature matching types (dtype('<U21'), dtype('<U21')) -> dtype('<U21')

我能够解决使用下面的列表理解。但我很好奇numpy是怎么工作的。你知道吗

def array_translate(array):

    intlist = [x for x in array if isinstance(x, int)]
    strlist = [x for x in array if isinstance(x, str)]
    return "".join(intlist*strlist for intlist, strlist in zip(intlist, strlist))


print(array_translate(["Cat", 2, "Dog", 3, "Mouse", 1]))    # => "CatCatDogDogDogMouse"

Answer 1

也许用重复的方法

z = array(['Cat', 'Dog', 'Mouse'], dtype='<U5')
"".join(np.repeat(z, (2, 3, 1)))
'CatCatDogDogDogMouse'

Answer 2

In [79]: arr = np.array(['Cat','Dog','Mouse'])                                  
In [80]: cnt = np.array([2,3,1])  

各种备选方案的时间安排。相对位置可能会随数组的大小而变化（以及是从列表还是数组开始）。所以你自己做测试：

In [93]: timeit ''.join(np.repeat(arr,cnt))                                     
7.98 µs ± 57.7 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

In [94]: timeit ''.join([str(wd)*i for wd,i in zip(arr,cnt)])                   
5.96 µs ± 167 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

In [95]: timeit ''.join(arr.astype(object)*cnt)                                 
13.3 µs ± 50.9 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

In [96]: timeit ''.join(np.char.multiply(arr,cnt))                              
27.4 µs ± 307 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)

In [100]: timeit ''.join(np.frompyfunc(lambda w,i: w*i,2,1)(arr,cnt))           
10.4 µs ± 164 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

In [101]: %%timeit f = np.frompyfunc(lambda w,i: w*i,2,1) 
     ...: ''.join(f(arr,cnt))                                                                       
7.95 µs ± 93.2 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

In [102]: %%timeit x=arr.tolist(); y=cnt.tolist() 
     ...: ''.join([str(wd)*i for wd,i in zip(x,y)])                                                                      
1.36 µs ± 39.7 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

np.repeat适用于所有类型的数组。你知道吗

列表理解使用字符串乘法，不应该被立即忽略。通常是最快的，尤其是从列表开始的时候。你知道吗

Object dtype将string dtype转换为Python字符串，然后将操作委托给string multiply。你知道吗

np.char将字符串方法应用于数组的元素。虽然方便，但很少快。你知道吗

编辑

In [104]: timeit ''.join(np.repeat(arr,cnt).tolist())                           
4.04 µs ± 197 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)