列表的乐趣

标签（空格分隔）： Python list

文章的原文是 Fun with Lists

1 删除元素

1.1 所有等于 X 值的元素

x = 4

a = [1, 2, 3, 4, 4, 5, 6, 1, 4]
for i in range(a.count(x)):
    a.pop(a.index(x))
print(a)

[1, 2, 3, 5, 6, 1]

a = [1, 2, 3, 4, 4, 5, 6, 1, 4]
b = [v for v in a if v != x]
print(b)
[1, 2, 3, 5, 6, 1]

print(a)
[1, 2, 3, 4, 4, 5, 6, 1, 4]

1.2 重复

警告!
该方法不负责列表项的顺序。

a = [1, 2, 3, 4, 3, 5, 1, 6]
b = list(set(a))
print(b)
[1, 2, 3, 4, 5, 6]

1.3 第一个元素

a = [1, 2, 3, 4, 5, 6]
b = a.pop(0)
print(b)
1

print(a)
[2, 3, 4, 5, 6]

a = [1, 2, 3, 4, 5, 6]
del a[0]
print(a)
[2, 3, 4, 5, 6]

a = [1, 2, 3, 4, 5, 6]
d = a[1:]
print(d)
[2, 3, 4, 5, 6]

print(a)
[1, 2, 3, 4, 5, 6]

1.4 最后一个元素

a = [1, 2, 3, 4, 5, 6]
b = a.pop()
print(b)
6

print(a)
[1, 2, 3, 4, 5]

a = [1, 2, 3, 4, 5, 6]
del a[-1]
print(a)
[1, 2, 3, 4, 5]

a = [1, 2, 3, 4, 5, 6]
c = a[:-1]
print(c)
[1, 2, 3, 4, 5]

print(a)
[1, 2, 3, 4, 5, 6]

1.5 第 n 个元素

n = 3
a = [1, 2, 3, 4, 5, 6]
b = a.pop(n - 1)
print(b)
3

print(a)
[1, 2, 4, 5, 6]

a = [1, 2, 3, 4, 5, 6]
del a[n - 1]
print(a)
[1, 2, 4, 5, 6]

a = [1, 2, 3, 4, 5, 6]
c = a[:n - 1] + a[n:]
print(c)
[1, 2, 4, 5, 6]

print(a)
[1, 2, 3, 4, 5, 6]

a = [1, 2, 3, 4, 5, 6]
c = [v for i, v in enumerate(a) if i != n - 1]
print(c)
[1, 2, 4, 5, 6]

print(a)
[1, 2, 3, 4, 5, 6]

2 替换元素

2.1 所有等于 x 值的元素

x = 4

a = [1, 2, 3, 4, 4, 5, 6, 1, 4]
for i in range(a.count(x)):
    a[a.index(x)] = 0
print(a)
[1, 2, 3, 0, 0, 5, 6, 1, 0]

a = [1, 2, 3, 4, 4, 5, 6, 1, 4]
b = [v if v != x else 0 for v in a ]
print(b)
[1, 2, 3, 0, 0, 5, 6, 1, 0]
print(a)
[1, 2, 3, 4, 4, 5, 6, 1, 4]

2.2 第一个元素

a = [1, 2, 3, 4]
a[0] = 0
print(a)
[0, 2, 3, 4]

a = [1, 2, 3, 4]
b = [0] + a[1:]
print(b)
[0, 2, 3, 4]
print(a)
[1, 2, 3, 4]

2.3 最后一个元素

a = [1, 2, 3, 4]
a[-1] = 0
print(a)
[1, 2, 3, 0]

a = [1, 2, 3, 4]
b = a[:-1] + [0]
print(b)
[1, 2, 3, 0]
print(a)
[1, 2, 3, 4]

2.4 第 n 个元素

n = 3

a = [1, 2, 3, 4]
a[n - 1] = 0
print(a)
[1, 2, 0, 4]

a = [1, 2, 3, 4]
b = [v if i != n -1 else 0 for i, v in enumerate(a)]
print(b)
[1, 2, 0, 4]
print(a)
[1, 2, 3, 4]

3 排序

3.1 按照字母顺序排列地 (不分大小写)

a = ['d', 'C', 'B', 'a']

b = sorted(a, key=lambda x: x.lower())
print(b)
['a', 'B', 'C', 'd']
print(a)
['d', 'C', 'B', 'a']

a.sort(key=lambda s: s.lower())
print(a)
['a', 'B', 'C', 'd']

3.2 按照字母顺序排列地 (分大小写)

a = ['d', 'C', 'B', 'a']

b = sorted(a)
print(b)
['B', 'C', 'a', 'd']
print(a)
['d', 'C', 'B', 'a']

a.sort()
print(a)
['B', 'C', 'a', 'd']

3.3 升序

a = ['a', 'c', 'd', 'b']

b = sorted(a, reverse=True)
print(b)
['d', 'c', 'b', 'a']
print(a)
['a', 'c', 'd', 'b']

a.sort(reverse=True)
print(a)
['d', 'c', 'b', 'a']

3.4 根据字符串长度

a = ['aaaa', 'B', 'CCC', 'dd']

b = sorted(a, key=lambda x: len(x))
print(b)
['B', 'dd', 'CCC', 'aaaa']
print(a)
['aaaa', 'B', 'CCC', 'dd']

a.sort(key=lambda x: len(x))
print(a)
['B', 'dd', 'CCC', 'aaaa']

4 其他

4.1 列表中的所有值相加

a = [1, 2.5, 7, 13221, 4.6545]
b = sum(a)
print(b)
13236.1545

4.2 在列表中新增元素

a = [1, 2, 3]
a.append(4)
print(a)
[1, 2, 3, 4]

4.3 使函数对列表中的每个元素生效

def func(x):
    print(x)

def func_2(x):
    return 2*x

a = [1, 2, 3, 4]
map(func, a)
1
2
3
4

b = map(func_2, a)
print(b)
[2, 4, 6, 8]
c = [func_2(x) for x in a]
print(c)
[2, 4, 6, 8]

4.4 两个列表的笛卡尔积 (矢量)

a = [1, 2, 3]
b = [4, 5, 6]

c = [(x, y) for x in a for y in b]
print(c)
[(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)]

import itertools
d = [p for p in itertools.product(a, b)]
print(d)
[(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)]

4.5 n 个列表的笛卡尔积 (矢量)

from pprint import pprint
import itertools

a = [[0, 1], [2, 3], [4, 5]]

b = [p for p in itertools.product(*a)]

print(b)
[(0, 2, 4),
 (0, 2, 5),
 (0, 3, 4),
 (0, 3, 5),
 (1, 2, 4),
 (1, 2, 5),
 (1, 3, 4),
 (1, 3, 5)]

4.6 检查两个列表是否有一个共同的元素

a = [1, 2, 0]
b = [3, 0, 4]

d = len((set(a) & set(b))) > 0
print(d)
True

4.7 检查一个列表是否包含值 x

a = [1, 2.5, 7, 13221, 4.6545]
if 7 in a:
    print('yes')
else:
    print('no')
yes

4.8 计算 x 出现在一个列表中的次数

a = [1, 2.5, 7, 13221, 4.6545, 7]
b = a.count(7)
print(b)
2

4.9 两个列表的不同

a = [1, 2, 3, 4]
b = [3, 4, 5, 6]

c = [x for x in a if not x in b]
print(c)
[1, 2]

d = list(set(a) - set(b))
print(d)
[1, 2]

e = list(set.difference(set(a), set(b)))
print(e)
[1, 2]

4.10 n 个列表的不同

a = [1, 2, 3, 4]
b = [[3, 5, 6, 7], [1, 8, 9, 10]]

a - b[0] - b[1]:

c = [x for x in a if not any([x in l for l in b])]
print(c)
[2, 4]

d = list(set(a).difference(*[set(l) for l in b]))
print(d)
[2, 4]

4.11 一个列表中首先出现的 n 个元素

n = 2

a = [1, 2, 3, 4, 5, 6]
b = a[:2]
print(b)
[1, 2]

4.12 一个列表中首先出现的 n 个非 x 值的元素

n = 3
x = 2

a = [1, 2, 2, 3, 2, 2, 2, 4, 5, 6]

gen = (v for v in a if v != x)
b = [gen.next() for i in range(n)]
print(b)
[1, 3, 4]
c = []
for v in a:
    if v != x:
        c.append(v)
    if len(c) == n:
        break
print(c)
[1, 3, 4]
4.13   Flatten a list of lists

a = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]

b = [i for s in a for i in s]
print(b)
[1, 2, 3, 4, 5, 6, 7, 8, 9]

4.14 把 x 插入 y 第一次出现之后的位置

x = 4
y = 3
a = [1, 2, 3, 5, 3, 6, 2]

try:
    a.insert(a.index(y) + 1, x)
except ValueError:
    a.append(x)
print(a)
[1, 2, 3, 4, 5, 3, 6, 2]
a = [1, 2, 3, 5, 3, 6, 2]
try:
    i = a.index(y)
    b = a[:i + 1] + [x] + a[i + 1:]
except ValueError:
    b = a + [x]
print(b)
[1, 2, 3, 4, 5, 3, 6, 2]
print(a)
[1, 2, 3, 5, 3, 6, 2]

4.15 把 x 插入 y 第一次出现之前的位置

x = 3
y = 4
a = [1, 2, 4, 5, 4, 6, 2]

try:
    a.insert(a.index(y), x)
except ValueError:
    a.append(x)
print(a)
[1, 2, 3, 4, 5, 4, 6, 2]

a = [1, 2, 4, 5, 4, 6, 2]
try:
    i = a.index(y)
    b = a[:i] + [x] + a[i:]
except ValueError:
    b = a + [x]
print(b)
[1, 2, 3, 4, 5, 4, 6, 2]
print(a)
[1, 2, 4, 5, 4, 6, 2]

4.16 两个列表的相同元素

a = [1, 2, 3, 4]
b = [3, 4, 5, 6]

c = [x for x in a if x in b]
print(c)
[3, 4]
d = list(set(a) & set(b))
print(d)
[3, 4]
e = list(set(a).intersection(b))
print(e)
[3, 4]

4.17 n 个列表的相同元素

a = [[1, 2, 3, 4], [3, 4, 5, 6], [1, 3, 7, 8]]

b = list(set.intersection(*[set(l) for l in a]))
print(b)
[3]
c = [x for x in a[0] if all([x in l for l in a[1:]])]
print(c)
[3]

4.18 Iterate over every other element of a list

a = [1, 2, 3, 4, 5, 6]
for x in a[::2]:
    print x
1
3
5

4.19 遍历列表的索引/值对

a = [1, 2.5, 7, 13221, 4.6545]
for i, v in enumerate(a):
    print('%i: %i' % (i, v))
0: 1
1: 2
2: 7
3: 13221
4: 4

4.20 遍历一个列表的元素

a = [1, 2, 3, 4]
for x in a:
    print x
1
2
3
4

4.21 列表中最大的数

a = [1, 2.5, 7, 13221, 4.6545]
b = max(a)
print(b)
13221
4.22   Last n elements of a list

n = 2

a = [1, 2, 3, 4, 5, 6]
b = a[-2:]
print(b)
[5, 6]

4.23 一个列表的长度

a = [1, 2, 3, 4]
l = len(a)
print(l)
4

4.24 合并两个列表

a = [1, 2, 3]
b = [4, 5, 6]

c = a + b
print(c)
[1, 2, 3, 4, 5, 6]
print(a)
[1, 2, 3]
print(b)
[4, 5, 6]

a.extend(b)
print(a)
[1, 2, 3, 4, 5, 6]
print(b)
[4, 5, 6]

4.25 列表的组合

from pprint import pprint
from itertools import permutations as perm

a = ['a', 'b', 'c']

b = list(perm(a))
pprint(b)
[('a', 'b', 'c'),
 ('a', 'c', 'b'),
 ('b', 'a', 'c'),
 ('b', 'c', 'a'),
 ('c', 'a', 'b'),
 ('c', 'b', 'a')]

c = [''.join(p) for p in perm(a)]
print(c)
['abc', 'acb', 'bac', 'bca', 'cab', 'cba']

d = [''.join(p) for p in perm(a, 2)]
print(d)
['ab', 'ac', 'ba', 'bc', 'ca', 'cb']

4.26 Prepend an element to a list

a = [2, 3, 4]

a.insert(0, 1)
print(a)
[1, 2, 3, 4]
a = [2, 3, 4]

b = [1] + a
print(b)
[1, 2, 3, 4]
print(a)
[2, 3, 4]

4.27 列表反序

a = [1, 2, 3, 4]
a.reverse()
print(a)
[4, 3, 2, 1]

a = [1, 2, 3, 4]
c = a[::-1]
print(c)
[4, 3, 2, 1]
print(a)
[1, 2, 3, 4]

4.28 列表中最小的数字

a = [1, 2.5, 7, 13221, 4.6545]
b = min(a)
print(b)
1

4.29 2 个列表的对称差

a = [1, 2, 3, 4]
b = [3, 4, 5, 6]

c = [x for x in a if x not in b] + [x for x in b if x not in a]
print(c)
[1, 2, 5, 6]

d = list(set(a) ^ set(b))
print(d)
[1, 2, 5, 6]

e = list(set(a).symmetric_difference(b))
print(e)
[1, 2, 5, 6]

4.30 n 个列表的对称差

a = [[1, 2, 3, 4], [3, 4, 5, 6], [1, 5, 7, 8]]

b = a[0]
for i in range(len(a) - 1):
    b = list(set(b) ^ set(a[i + 1]))
print(b)
[2, 6, 7, 8]
tmp = {}
for v in [i for s in a for i in s]:
    tmp[v] = (v not in tmp)
c = [k for k, v in tmp.iteritems() if v]
print(c)
[2, 6, 7, 8]

4.31 两个列表的交集

a = [1, 2, 3, 4]
b = [3, 4, 5, 6]

c = [x for x in set(a + b)]
print(c)
[1, 2, 3, 4, 5, 6]
d = list(set(a) | set(b))
print(d)
[1, 2, 3, 4, 5, 6]
e = list(set(a).union(b))
print(e)
[1, 2, 3, 4, 5, 6]

4.32 n 个列表的交集

a = [[1, 2, 3, 4], [3, 4, 5, 6], [1, 6, 7, 8]]

b = list(set.union(*[set(l) for l in a]))
print(b)
[1, 2, 3, 4, 5, 6, 7, 8]
import itertools
c = [x for x in set(itertools.chain(*a))]
print(c)
[1, 2, 3, 4, 5, 6, 7, 8]

4.33 列表的左边加 0

max_length = 10

a = [1, 2, 3, 4]

b = [0]*(max_length - len(a)) + a
print(b)
[0, 0, 0, 0, 0, 0, 1, 2, 3, 4]

4.34 列表的右边加 0

max_length = 10

a = [1, 2, 3, 4]

b = a + [0]*(max_length - len(a))
print(b)
[1, 2, 3, 4, 0, 0, 0, 0, 0, 0]