检查字符串是否以XXXX开头

我想知道如何在Python中检查字符串是否以“hello”开头。

在Bash中，我通常这样做:

if [[ "$string" =~ ^hello ]]; then
do something here
fi

如何在Python中实现同样的功能?

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小开
最佳答案 
aString = "hello world"
aString.startswith("hello")

关于startswith的更多信息。

小开

RanRag已经回答了它为你的具体问题。

然而，更一般地说，你在做什么

if [[ "$string" =~ ^hello ]]

是正则表达式匹配。要在Python中执行相同的操作，您将执行:

import re
if re.match(r'^hello', somestring):
# do stuff

显然，在这种情况下，somestring.startswith('hello')更好。

小开

也可以这样做..

regex=re.compile('^hello')


## THIS WAY YOU CAN CHECK FOR MULTIPLE STRINGS
## LIKE
## regex=re.compile('^hello|^john|^world')


if re.match(regex, somestring):
print("Yes")
小开

如果你想将多个单词与你的魔法单词匹配，你可以将要匹配的单词作为元组传递:

>>> magicWord = 'zzzTest'
>>> magicWord.startswith(('zzz', 'yyy', 'rrr'))
True

startswith接受字符串或字符串元组。

小开

我做了一个小实验，看看是哪种方法

  • string.startswith('hello')
  • string.rfind('hello') == 0
  • string.rpartition('hello')[0] == ''
  • string.rindex('hello') == 0

最有效的方法是返回某个字符串是否以另一个字符串开头。

下面是我所做的许多测试运行之一的结果，其中每个列表的顺序是显示在我使用的while循环的每次迭代中，(以秒为单位)解析500万个以上每个表达式所花费的最少时间:

['startswith: 1.37', 'rpartition: 1.38', 'rfind: 1.62', 'rindex: 1.62']
['startswith: 1.28', 'rpartition: 1.44', 'rindex: 1.67', 'rfind: 1.68']
['startswith: 1.29', 'rpartition: 1.42', 'rindex: 1.63', 'rfind: 1.64']
['startswith: 1.28', 'rpartition: 1.43', 'rindex: 1.61', 'rfind: 1.62']
['rpartition: 1.48', 'startswith: 1.48', 'rfind: 1.62', 'rindex: 1.67']
['startswith: 1.34', 'rpartition: 1.43', 'rfind: 1.64', 'rindex: 1.64']
['startswith: 1.36', 'rpartition: 1.44', 'rindex: 1.61', 'rfind: 1.63']
['startswith: 1.29', 'rpartition: 1.37', 'rindex: 1.64', 'rfind: 1.67']
['startswith: 1.34', 'rpartition: 1.44', 'rfind: 1.66', 'rindex: 1.68']
['startswith: 1.44', 'rpartition: 1.41', 'rindex: 1.61', 'rfind: 2.24']
['startswith: 1.34', 'rpartition: 1.45', 'rindex: 1.62', 'rfind: 1.67']
['startswith: 1.34', 'rpartition: 1.38', 'rindex: 1.67', 'rfind: 1.74']
['rpartition: 1.37', 'startswith: 1.38', 'rfind: 1.61', 'rindex: 1.64']
['startswith: 1.32', 'rpartition: 1.39', 'rfind: 1.64', 'rindex: 1.61']
['rpartition: 1.35', 'startswith: 1.36', 'rfind: 1.63', 'rindex: 1.67']
['startswith: 1.29', 'rpartition: 1.36', 'rfind: 1.65', 'rindex: 1.84']
['startswith: 1.41', 'rpartition: 1.44', 'rfind: 1.63', 'rindex: 1.71']
['startswith: 1.34', 'rpartition: 1.46', 'rindex: 1.66', 'rfind: 1.74']
['startswith: 1.32', 'rpartition: 1.46', 'rfind: 1.64', 'rindex: 1.74']
['startswith: 1.38', 'rpartition: 1.48', 'rfind: 1.68', 'rindex: 1.68']
['startswith: 1.35', 'rpartition: 1.42', 'rfind: 1.63', 'rindex: 1.68']
['startswith: 1.32', 'rpartition: 1.46', 'rfind: 1.65', 'rindex: 1.75']
['startswith: 1.37', 'rpartition: 1.46', 'rfind: 1.74', 'rindex: 1.75']
['startswith: 1.31', 'rpartition: 1.48', 'rfind: 1.67', 'rindex: 1.74']
['startswith: 1.44', 'rpartition: 1.46', 'rindex: 1.69', 'rfind: 1.74']
['startswith: 1.44', 'rpartition: 1.42', 'rfind: 1.65', 'rindex: 1.65']
['startswith: 1.36', 'rpartition: 1.44', 'rfind: 1.64', 'rindex: 1.74']
['startswith: 1.34', 'rpartition: 1.46', 'rfind: 1.61', 'rindex: 1.74']
['startswith: 1.35', 'rpartition: 1.56', 'rfind: 1.68', 'rindex: 1.69']
['startswith: 1.32', 'rpartition: 1.48', 'rindex: 1.64', 'rfind: 1.65']
['startswith: 1.28', 'rpartition: 1.43', 'rfind: 1.59', 'rindex: 1.66']

我相信从一开始就很明显startswith方法是最有效的，因为返回字符串是否以指定的字符串开头是它的主要目的。

令我惊讶的是，看似不切实际的string.rpartition('hello')[0] == ''方法总是能找到一种方法，在string.startswith('hello')方法之前被列在前面，时不时地。结果表明，使用str.partition来确定一个字符串是否以另一个字符串开头，比同时使用rfind和rindex更有效。

我注意到的另一件事是string.rfind('hello') == 0和string.rindex('hello') == 0正在进行一场很好的战斗，每个都从第四名上升到第三名，从第三名下降到第四名，这是有道理的，因为它们的主要目的是相同的。

代码如下:

from time import perf_counter


string = 'hello world'
places = dict()


while True:
start = perf_counter()
for _ in range(5000000):
string.startswith('hello')
end = perf_counter()
places['startswith'] = round(end - start, 2)


start = perf_counter()
for _ in range(5000000):
string.rfind('hello') == 0
end = perf_counter()
places['rfind'] = round(end - start, 2)


start = perf_counter()
for _ in range(5000000):
string.rpartition('hello')[0] == ''
end = perf_counter()
places['rpartition'] = round(end - start, 2)


start = perf_counter()
for _ in range(5000000):
string.rindex('hello') == 0
end = perf_counter()
places['rindex'] = round(end - start, 2)
    

print([f'{b}: {str(a).ljust(4, "4")}' for a, b in sorted(i[::-1] for i in places.items())])

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