Question

如果字符串中出现一次，则以下代码显示“word”的位置。如何更改我的代码，以便如果“word”在字符串中出现多次，它将打印所有位置？

string = input("Please input a sentence: ")
word = input("Please input a word: ")
string.lower()
word.lower()
list1 = string.split(' ')
position = list1.index(word)
location = (position+1)
print("You're word, {0}, is in position {1}".format (word, location))

Answer 1

使用enumerate：

[i for i, w in enumerate(s.split()) if w == 'test']

示例：

s = 'test test something something test'

输出：

[0, 1, 4]

但我想这不是你想要的，如果你需要为字符串中的单词开始索引我建议使用re.finditer：

import re

[w.start() for w in re.finditer('test', s)]

同一s的输出将是：

[0, 5, 30]

Answer 2

sentence = input("Please input a sentence: ")
word = input("Please input a word: ")
sentence = sentence.lower()
word = word.lower()
wordlist = sentence.split(' ')
print ("Your word '%s' is in these positions:" % word)
for position,w in enumerate(wordlist):
    if w == word:
        print("%d" % position + 1)

Answer 3

另一种不会在空间上分裂的解决方案。

def multipos(string, pattern):

    res = []
    count = 0
    while True:
        pos = string.find(pattern)
        if pos == -1:
            break
        else:
            res += [pos+count]
            count += pos+1
            string = string[pos+1:]

    return res


test = "aaaa 123 bbbb 123 cccc 123"
res = multipos("aaaa 123 bbbb 123 cccc 123", "123")
print res
for a in res:
    print test[a:a+3]

脚本输出：

% python multipos.py
[5, 14, 23]
123
123
123

如何在字符串中找到多于一个子字符串的位置（Python 3.4.3 shell）

3 个答案: