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 [[ "$string" =~ ^hello ]]; then
do something here
fi
如何在Python中实现同样的功能?
我想知道如何在Python中检查字符串是否以“hello”开头。
在Bash中,我通常这样做:
if [[ "$string" =~ ^hello ]]; then
do something here
fi
如何在Python中实现同样的功能?
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。
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")
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。
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())])