检查字符串是否以XXXX开头

Question

我想知道如何在Python中检查字符串是否以"hello"开头.

在Bash我经常做:

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

我如何在Python中实现相同的目标？

Answer 1

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

Answer 2

RanRag已经针对您的具体问题回答了这个问题.

但是,更一般地说,你在做什么

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

是一个正则表达式匹配.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')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

查看文档.

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

我做了一个小实验来看看这些方法中的哪一个

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

返回某个字符串是否以另一个字符串开头是最有效的。

这是我进行的多次测试运行之一的结果，其中每个列表都按顺序显示在我使用的循环的每次迭代期间解析上述每个表达式 500 万个所需的最短时间（以秒为单位）while :

['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该startswith方法将是最有效的，因为返回字符串是否以指定字符串开头是其主要目的。

让我惊讶的是，看似不切实际的string.rpartition('hello')[0] == ''方法总是时不时地找到一种方法先列出来，列在string.startswith('hello')方法之前。结果表明，使用来确定一个字符串是否以另一个字符串开头比同时使用和str.partition更有效。rfindrindex

我注意到的另一件事是，string.rfind('hello') == 0并且string.rindex('hello') == 0正在进行一场精彩的战斗，每个人都从第四名上升到第三名，然后从第三名下降到第四名，这是有道理的，因为他们的主要目的是相同的。

这是代码：

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())])