如何在python中的数组元素中查找特定字符串

Question

我在 python 中有一个字符串列表，如果列表中的一个元素包含单词“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”，因此它应该打印消息。

Answer 1

如果您想使用正则表达式执行此操作，则不能使用 in 运算符。请改用re.search()。但它适用于 字符串， 而不是整个列表。

for elt in a:
    if re.search(my_regexp, elt):
        print "Matched"
        break # stop looking

或者更实用的风格：

if any(re.search(my_regexp, elt) for elt in a)):
    print "Matched"

Answer 2

您不需要正则表达式，只需使用any。

>>> a = ["paul Parthipan","paul","sdds","sdsdd"]
>>> last_name = "Parthipan".lower()
>>> if any(last_name in name.lower() for name in a):
...     print("Matched")
... 
Matched

Answer 3

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"

这部分还有什么用：

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

编辑：

我太盲目了，看不到你在这里需要正则表达式做什么。如果你能给出一些真实的输入和输出，那将是最好的。这是一个小例子，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"

但如果真的需要正则表达式部分，我无法想象如何在“Parthipan”中找到“(?mis)Parthipan”。最简单的是在“（？mis）Parthipan”中反向“Parthipan”。喜欢这里...

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：

嗯，用正则表达式你几乎没有可能......

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，只有第一个似乎没问题，所以只有 re.search 给出了正确的解决方案：

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