python - 如何在python中的数组元素中找到特定的字符串

我在python中有一个字符串列表，如果列表中的元素包含单词“ parthipan”，我应该打印一条消息。但是下面的脚本不起作用

import re
a = ["paul Parthipan","paul","sdds","sdsdd"]
last_name = "Parthipan"
my_regex = r"(?mis){0}".format(re.escape(last_name))
if my_regex in a:
    print "matched"

列表的第一个元素包含单词“ parthipan”，因此应打印该消息。

最佳答案

为什么不：

a = ["paul Parthipan","paul","sdds","sdsdd"]
last_name = "Parthipan"
if any(last_name in ai for ai in a):
    print "matched"

这部分又是什么：

...
import re
my_regex = r"(?mis){0}".format(re.escape(last_name))
...

编辑：

我实在太盲目了，看不明白你在做什么。最好提供一些实际的输入和输出。这是一个小的示例，也可以通过这种方式完成：

a = ["paul Parthipan","paul","sdds","sdsdd",'Mala_Koala','Czarna,Pala']
last_name = "Parthipan"
names=[]
breakers=[' ','_',',']
for ai in a:
    for b in breakers:
        if b in ai:
            names.append(ai.split(b))
full_names=[ai for ai in names if len(ai)==2]
last_names=[ai[1] for ai in full_names]

if any(last_name in ai for ai in last_names):
    print "matched"

但是如果真的需要正则表达式，我无法想象如何在“ Parthipan”中找到“（？mis）Parthipan”。最简单的是在“（？mis）Parthipan”中使用相反的方向“ Parthipan”。像这儿...

import re
a = ["paul Parthipan","paul","sdds","sdsdd",'Mala_Koala','Czarna,Pala']
last_name = "Parthipan"
names=[]
breakers=[' ','_',',']
for ai in a:
    for b in breakers:
        if b in ai:
            names.append(ai.split(b))
full_names=[ai for ai in names if len(ai)==2]
last_names=[r"(?mis){0}".format(re.escape(ai[1])) for ai in full_names]
print last_names
if any(last_name in ai for ai in last_names):
    print "matched"

编辑：

是的，使用正则表达式，您几乎没有...

import re
a = ["paul Parthipan","paul","sdds","sdsdd",'jony-Parthipan','koala_Parthipan','Parthipan']
lastName = "Parthipan"
myRegex = r"(?mis){0}".format(re.escape(lastName))

strA=';'.join(a)
se = re.search(myRegex, strA)
ma = re.match(myRegex, strA)
fa = re.findall(myRegex, strA)
fi=[i.group() for i in re.finditer(myRegex, strA, flags=0)]
se = '' if se is None else se.group()
ma = '' if ma is None else ma.group()

print se, 'match' if any(se) else 'no match'
print ma, 'match' if any(ma) else 'no match'
print fa, 'match' if any(fa) else 'no match'
print fi, 'match' if any(fi) else 'no match'

输出，只有第一个看起来可以，因此只有re.search可以提供适当的解决方案：

Parthipan match
 no match
['Parthipan', 'Parthipan', 'Parthipan', 'Parthipan'] match
['Parthipan', 'Parthipan', 'Parthipan', 'Parthipan'] match