按元素比较字符串1-d numpy数组 [英] comparing a string 1-d numpy array elementwise

问题描述

我有两个包含字符串的1-d数组(a和b)，我想按元素进行比较以获取输出c，如下所示.我尝试将其转换为设置值并进行比较，但是并没有给出正确的解决方案.同样，logical_xor不适用于字符串.我可以编写一个循环来执行此操作，但是这样做却违背了使用数组的目的，没有循环的最佳方法是什么?

I have two 1-d arrays (a and b) containing strings, which I want to compare element wise to get output c like shown below. I tried converting it to set and comparing, however that does not give the correct solution. Also logical_xor does not work for string. I can write a loop to do this but then it defeats the purpose of using arrays, What can be the best way to do this without a loop?

  >>  a
      array(['S', 'S', 'D', 'S', 'N', 'S', 'A', 'S', 'M'], 
          dtype='|S1')
  >>  b
      array(['T', 'I', 'D', 'N', 'G', 'B', 'A', 'J', 'M'], 
          dtype='|S1')

  >> c 
     array([False, False, True, False, False, False, True, False, True], 
      dtype=bool)

推荐答案

只需使用ndarray的__eq__方法，即==

Just use the ndarray's __eq__ method, i.e. ==

>>> a = array(['S', 'S', 'D', 'S', 'N', 'S', 'A', 'S', 'M'], dtype='|S1')
>>> b = array(['T', 'I', 'D', 'N', 'G', 'B', 'A', 'J', 'M'], dtype='|S1')
>>> a == b
array([False, False,  True, False, False, False,  True, False,  True], dtype=bool)

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