您不需要加入包含字符串的电子邮件的初始数组。

<version>

提取每个列表项中的电子邮件：

import re

emails = ['John Kennedy <jk123@gmail.com> or <johnk123@hotmail.com>','Adam Hartley <ah123@yahoo.com>','Ben Saunders <benji@live.co.uk>']
def myfunction(bigstring):
    result = []
    for s in bigstring:
        result.append(re.findall(r'[\w.-]+@[\w.-]+',s))
    return result

print(myfunction(emails))
# => [['jk123@gmail.com','johnk123@hotmail.com'],['ah123@yahoo.com'],['benji@live.co.uk']]

请参见Python demo。

实际上，bigstring在这里是一个不好的名字，因为它是一个列表。考虑重命名为my_list之类的东西。

对于正则表达式，您无需在字符类中转义点。

不要将整个字符串连接到函数中，并像这样通过迭代逐一传递字符串

import re
def myfunction(bigstring):
    return  re.findall(r'[\w\.-]+@[\w\.-]+',bigstring)

emails = ['John Kennedy <jk123@gmail.com> or <johnk123@hotmail.com>','Ben Saunders <benji@live.co.uk>']


output = []
for emailstring in emails:
    output.append((myfunction(emailstring)))
print(output)

使用列表理解

output = [ myfunction(email) for email in emails ]
print(output)

使用地图

print(map(myfunction,emails))

输出

[['jk123@gmail.com',['benji@live.co.uk']]

def myfunction(bigstring):
    #str = ' '.join(bigstring)
    #my_string = str
    str1 = re.findall(r'[\w\.-]+@[\w\.-]+',bigstring)
    return str1

import re
output=[]
emails = ['John Kennedy <jk123@gmail.com> or <johnk123@hotmail.com>','Ben Saunders <benji@live.co.uk>']
for item in emails:
    output.append(myfunction(item))

print(output)

    def myfunction(bstr):
        #str = ' '.join(bstr)
        #my_string = str
        str1 = re.findall(r'[\w\.-]+@[\w\.-]+',bstr)
        return str1
    import re
    output=[]
    emails = ['John Kennedy <jk123@gmail.com> or <johnk123@hotmail.com>','Adam 
              Hartley <ah123@yahoo.com>','Ben Saunders <benji@live.co.uk>']
     for item in emails:
        output.append(myfunction(item))