[英]Python 2.7: how to find the index of the first letter of each word in a string
[英]find not just the first index of a substring in a string - python 2.7
所以我知道str.index(substring,begin,end = len(str))返回從begin開始的子字符串的第一個索引。 獲取字符串的下一個索引的方法是否更好(更快,更清晰),而不是簡單地將開始索引更改為最后一次出現的索引+目標字符串的長度? 即(這是我正在運行的代碼)
full_string = "the thing is the thingthe thing that was the thing that did something to the thing."
target_string = "the thing"
count = full_string.count(target_string)
print 'Count:', count
indexes = []
if (count > 0):
indexes.append(full_string.index(target_string))
i = 1
while (i < count):
start_index = indexes[len(indexes) - 1] + len(target_string)
current_index = full_string.index(target_string, start_index)
indexes.append(current_index)
i = i + 1
print 'Indexes:', indexes
輸出:
Count: 5
Indexes: [0, 13, 22, 41, 73]
您可以使用re.finditer
和列表理解:
>>> import re
>>> [m.start() for m in re.finditer(target_string, full_string)]
[0, 13, 22, 41, 73]
匹配對象有兩個有用的方法.start()
和.end()
,它們返回當前組匹配的子字符串的開始和結束索引。
使用切片的另一種方法:
>>> [i for i in xrange(len(full_string) - len(target_string) + 1)
if full_string[i:i+len(target_string)] == target_string]
[0, 13, 22, 41, 73]
您可以創建一個簡單的生成器:
def gsubstrings(string, sub):
i = string.find(sub)
while i >= 0:
yield i
i = string.find(sub, len(sub) + i)
>>> list(gsubstrings(full_string, target_string))
[0, 13, 22, 41, 73]
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