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")

关于你的具体问题，RanRag已经回答过了。

然而，更一般地说，你在做什么

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

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. startwith ('hello')更好。

如果你想匹配多个单词到你的魔法单词，你可以将单词作为一个元组传递给匹配:

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

Startswith接受一个字符串或字符串元组。

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")

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

更多关于startwith的信息。

我做了一个小实验，看看是哪种方法

string.startswith(“你好”) 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显示了在我使用的每次while循环迭代中解析500万个以上每个表达式所花费的最少时间(以秒为单位):

['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。

让我惊讶的是，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。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有一个很好的战斗正在进行，每个从第四上升到第三，从第三下降到第四，这是有意义的，因为他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())])