Question

我想知道如何在Python中检查字符串是否以“hello”开头。

在Bash我经常这样做：

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

如何在Python中实现相同的目标？

Answer 1

aString = "hello world"
aString.startswith("hello")

有关startwith

的更多信息

Answer 2

RanRag has already answered针对您的具体问题。

然而，更一般地说，你正在做什么

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

是正则表达式匹配。要在Python中执行相同的操作，您可以：

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

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

Answer 3

如果你想将多个单词与你的魔法单词匹配，你可以传递单词作为元组匹配：

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

注意：startswith需要str or a tuple of str

请参阅docs。

Answer 4

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")

Answer 5

我做了一个小实验，看看其中哪种方法

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

最有效地返回某个字符串是否以另一个字符串开头的信息。

这是我进行的许多测试运行之一的结果，其中每个列表的排列顺序显示了（以秒为单位）解析每一次的500万次所需的最少时间在我使用的while循环的每次迭代过程中，上述表达式均如此：

['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其主要目的。

令我惊讶的是，看似不切实际的string.rpartition('hello')[0] == ''方法总是时不时地找到一种在string.startswith('hello')方法之前首先被列出的方法。结果表明，与同时使用str.partition和rfind相比，使用rindex确定一个字符串是否以另一个字符串开头更有效。

我注意到的另一件事是string.rindex('hello') == 0和string.rindex('hello') == 0的战斗正在进行中，它们分别从第四位上升到第三位，而从第三位下降到第四位，这很有意义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())])

检查字符串是否以XXXX开头

5 个答案: