检查字符串是否以XXXX开头

Question

我想知道如何在Python中检查一个字符串是否以"hello"开头。在Bash中，我通常会这样做：

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

我该如何在Python中实现相同的功能？

- John Marston

Answer 1

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

关于startswith的更多信息。

- RanRag

Answer 2

RanRag已经回答过您的具体问题。

但是，更一般地说，您所做的是使用

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

判断一个正则表达式是否匹配。在Python中，你可以这样做:

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

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

- Shawabawa

---

只是想补充一点，就我所做的事情而言，re.match和re.sub始终比任何其他方法慢得多。 - Michał Leon

Answer 3

如果您想将多个单词与您的魔术单词匹配，可以将要匹配的单词作为元组传递：

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

startswith函数接受一个字符串或字符串元组。

- user1767754

Answer 4

也可以用这种方式完成。

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")

- Aseem Yadav

Answer 5

我进行了一个小实验，看一下以下方法中哪个最有效率地返回一个字符串是否以另一个字符串开头：

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

这是我进行多次测试之一的结果，每次测试循环使用while，并对每个表达式的500万次执行时间进行排序，以显示所需的最短时间（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，因为它的主要目的是返回一个字符串是否以指定字符串开头。

令人惊讶的是，看似不切实际的 string.rpartition('hello')[0] == '' 方法总能在 string.startswith('hello') 方法之前排名第一。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 更有效率。

另外我注意到，string.rfind('hello') == 0 和 string.rindex('hello') == 0 之间有一场激烈的争夺，它们时而从第四上升到第三，时而从第三下降到第四，这很有意义，因为它们的主要目的是相同的。

以下是代码：

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())])

- Ann Zen