文章目录

• 嵌套列表推导式
• 函数做对象
• • 将多个函数作为对象放到一个列表中
  • 将函数作为对象传入 map 中对列表元素批量操作

嵌套列表推导式

name_lsts = [["jee","peer","perry"]
             ,["darl","cristin","wang"]]
all_names = [name for name_lst in name_lsts for name in name_lst]
all_names

['jee', 'peer', 'perry', 'darl', 'cristin', 'wang']

# 相当于
all_names = []
for name_lst in name_lsts:
    for name in name_lst:
        all_names.append(name)
        
all_names

['jee', 'peer', 'perry', 'darl', 'cristin', 'wang']

all_names = [name for name_lst in name_lsts for name in name_lst if name.count("e")>=2]
all_names

['jee', 'peer']

# 相当于
all_names = []
for name_lst in name_lsts:
    for name in name_lst:
        if name.count("e") >= 2:
            all_names.append(name)
        
all_names

['jee', 'peer']

函数做对象

将多个函数作为对象放到一个列表中

import re

strs = ["#hello","daf ?","qip! "]

def remove_punctuation(str):
    return re.sub('[?!# ]','',str)
    
# 把这些处理的函数操作当做对象放在列表中，然后依次调取对数据进行处理
operators = [str.strip, remove_punctuation, str.title]

for i in range(len(strs)):
    for operator in operators:
        strs[i] = operator(strs[i])
print(strs)

['Hello', 'Daf', 'Qip']

strs = ["#hello","daf ?","qip! "]
operators = [str.strip, remove_punctuation, str.title]

def tackle_strs(strs,operators):
    for i in range(len(strs)):
        for operator in operators:
            strs[i] = operator(strs[i])

tackle_strs(strs,operators)

strs

['Hello', 'Daf', 'Qip']

将函数作为对象传入 map 中对列表元素批量操作

strs = ["#hello","daf ?","qip! "]

# 使用 map 将某个函数应用到某个列表上
[*map(str.strip,strs)]

['#hello', 'daf ?', 'qip!']