python之列表list方法分类整理

题记

将自带的方法分类整理、验证并做成例子, 方便记忆与理解

起初

是为了方便自己查找, 后来因为越来越受欢迎, 所以分享出来

基本的增删改查，insert,pop

def test_1():
# 增删改查，增和删
list1 = ['a', 'b', 'c']
list1.append('d')
assert ['a', 'b', 'c', 'd'] == list1, 'append error'
del list1[0]
assert ['b', 'c', 'd'] == list1, 'del error'
list1[0] = 'e'
assert ['e', 'c', 'd'] == list1, 'update error'
c = list1[1]
assert 'c' == c, 'query error'

list1.insert(1, 'f')
assert ['e', 'f', 'c', 'd'] == list1, 'insert error'
list1.pop(1)
assert ['e', 'c', 'd'] == list1, 'pop error'
list1.pop()
assert ['e', 'c'] == list1, 'pop 2 error'

2.list方法, +操作符, in, enumerate

def test_2():
# 方法：list(p), + , in,enumerate
list1 = list('abc')
assert ['a', 'b', 'c'] == list1, 'list(p) 方法错误'
list1 = list1 + ['d', 'e']
assert ['a', 'b', 'c', 'd', 'e'] == list1, 'list相加错误'
assert 'a' in list1, 'list的in操作错误'

list2 = list({'name': 'a', 'age': 19})
assert ['name', 'age'] == list2, 'list(p) 方法错误'

indexs = [i for i, _ in enumerate(list2)]
assert [0, 1] == indexs, 'enumerate error'

3. append和extend

def test_3():
# append 和 extend
list1 = ['a', 'b', 'c']
list2 = ['d', 'e']
list_append = list1.copy()
list_append.append(list2)
assert ['a', 'b', 'c', ['d', 'e']] == list_append, 'list append方法错误'
list_extend = list1.copy()
list_extend.extend(list2)
assert ['a', 'b', 'c', 'd', 'e'] == list_extend, 'list extentd 方法错误'

4.index 和 sort方法

def test_4():
# list.index(obj),sort
list1 = ['b', 'c', 'a']
i = 0
for v in list1:
assert list1.index(v) == i, f'list index方法错误 i:{i},v:{v}'
i += 1

list1.sort()
assert ['a', 'b', 'c'] == list1, 'list sort 方法错误'
list1.sort(reverse=True)
assert ['c', 'b', 'a'] == list1, 'list sort reverse 方法错误'

list2 = [{
'name': 'b',
'age': 20
}, {
'name': 'a',
'age': 18
}, {
'name': 'b',
'age': 19
}]
list2.sort(key=lambda kv: kv['name'])
assert {'name': 'a', 'age': 18} == list2[0], 'list sort key方法错误'
list2.sort(key=lambda kv: kv['age'])
assert [{
'name': 'a',
'age': 18
}, {
'name': 'b',
'age': 19
}, {
'name': 'b',
'age': 20
}] == list2, 'list sort key 方法错误'
list2.sort(key=lambda kv: (kv['name'], kv['age']))
assert [{
'name': 'a',
'age': 18
}, {
'name': 'b',
'age': 19
}, {
'name': 'b',
'age': 20
}] == list2, 'list sort key 方法错误'

5.浅复制和深度复制

import copy

def test_5():
# 浅复制 和 深度复制
# 只有list1中的数据是引用类型的时候，才会考虑deepcopy，否则，平常的copy就行
# 也就是说,如果只是修改a,b,c, list3,list4,list5 都不会影响list1
list1 = ['a', 'b', 'c', ['e', 'f']]
list2 = list1
list3 = list1[:]
list4 = list1.copy()
list5 = [i for i in list1]
list6 = copy.deepcopy(list1)

assert id(list1) == id(list2), '== error'
assert id(list1) != id(list3), '[:] error'
assert id(list1) != id(list4), 'list copy error'
assert id(list1) != id(list5), 'for error'
assert id(list1) != id(list6), 'deepcopy error'

list6[3][0] = 'd'
assert ['a', 'b', 'c', ['e', 'f']] == list1, 'deepcopy 1 error'
assert ['a', 'b', 'c', ['d', 'f']] == list6, 'deepcopy 2 error'

6.倒着删除: 为了防止删除时报错或者数据不对

# 正确的删除方式
def test_6():
list1 = ['a', 'b', 'c', 'd']
for i in range(len(list1) - 1, -1, -1):
del list1[i]
assert [] == list1, 'list delete error'
# 错误的方式
def test_7():
# 不要这样删除:https://blog.51cto.com/u_14246112/3157689
list1 = ['a', 'b', 'c']
for i in list1:
list1.remove(i)
# 本意是删除所有元素,但是删除的过程中列表发生了位移,所以删除了 a和c
assert ['b'] == list1, 'list delete 2 error'

7.拆分: n:小集合的长度为n

def test_8():
a1 = [i for i in range(100)]
s1 = split_list_by_size(a1, 7)
# 每个小集合应该都是7,最后一个是小于等于7
assert len(s1[-1]) <= 7, 'split size 1 error'
s1.pop()
for i in s1:
assert len(i) == 7, 'split size 2 error'

def split_list_by_size(raw_list, n):
# n:小集合的长度为n
l = len(raw_list)
return [raw_list[i:i + n] for i in range(0, l, n)]
8. 拆分: n:有多少个小集合

def test_9():
a1 = [i for i in range(100)]
s1 = split_list_by_number(a1, 7)
# 大集合中有7个小集
assert len(s1) == 7, 'split number 2 error'

def split_list_by_number(raw_list, n):
# n:有多少个小集合
l = len(raw_list)
s = int(l / n) + 1
return [raw_list[i:i + s] for i in range(0, l, s)]


审核编辑：刘清