找到字符串中每个不同字符的所有位置的最快方法

Question

Suppose we have a string, say, "122113" and we are supposed to find all the occurrences of every character in the String.

天真的方法是这样的：

string = str( raw_input() )  # for example: "122113"
distinct_char = list( set(string) )
occurrences=[]
for element in distinct_char:
    temp=[]
    for j in range(len(string)):
        if(string[j]==element):
            temp.append(j)
    occurrences.append(temp)
print(occurrences)  # output for "122113" would be [[0, 3, 4], [1, 2], [5]]
                    #because 1 occurrs at : 0, 3, 4
                    #        2 occurrs at : 1, 2
                    #        3 occurrs at : 5

But, this is quite slow if the length of the String is Large. So, is there any faster solution?

（假设字符串仅由较低的英文字母组成，字符串的长度可能为$10^12$

Answer 1

一种可能的解决方案是将字符串转换为数字，并使用数字来增加数组中的值。代码可以如下：

import numpy as np

def alph_to_num(alph):
    return ord(alph.lower())-97

string='alsblnasdglasdaoerngeaglbneronbiwernblnerl'
array=np.zeros(26)

for alph in string:
    index=alph_to_num(alph)
    array[index]=array[index]+1
print(array)

它给出：[5. 4. 0. 2. 5. 0. 3. 0. 1. 0. 0. 6. 0. 6. 2. 0. 0. 4. 3. 0. 0. 0. 1. 0. 0. 0.]

这里我已经做了长度为26的数组，因为你知道它只是小写英文字母。这也意味着更容易解释结果列表。你知道吗