List Comprehension - A Short Guide¶
难度: 3
时长: 30 min
Comprehension 用途¶
对（列表中的）每个成页应用相同的操作
对 每个成员应用同一种操作¶
问题： 将 names 中的每个名字改为大写¶
names = ['alice', 'tom', 'bob']
solution 1: (for in)¶
upper_names = []
for name in names:
upper_names.append(name.upper())
solution 2: map¶
upper_names = list(map(str.upper, names))
upper_names = list(map(lambda x: x.upper(), names))
solution 3: comprehension¶
upper_names = [name.upper() for name in names]
upper_names = (name.upper() for name in names) # iterable
upper_names = list(upper_names)
模式总结与语法¶
输出是 List，且针对成员做处理，完了之后，形成一个新的 List。
newlist = [expression for item in iterable_list
练习： 如果是想把所有名字的字母逆序，该怎么实现？试一试
条件过滤¶
问题： 将包含了字符 'a' 的 list 成员找出来，形成一个新的 List:¶
fruits = ["apple", "banana", "cherry", "kiwi", "mango"]
newlist = []
for x in fruits:
if "a" in x:
newlist.append(x)
if¶
这符合典型的 List Comprehension 特征，但需要加一个 if 条件：
fruits = ["apple", "banana", "cherry", "kiwi", "mango"]
newlist = [x for x in fruits if "a" in x]
模式总结与语法¶
newlist = [expression for item in iterable if condition == True]
The return value is a new list, leaving the old list unchanged.
The condition is like a filter that only accepts the items that valuate to True.
Examples¶
不作任何处理，原样复制¶
newlist = [x for x in range(10)]
对全体成员转大写¶
newlist = [x.upper() for x in fruits]
filtering exclude | 排除¶
Only accept items that are not "apple" （filtering）:
newlist = [x for x in fruits if x != "apple"]
真假两种情况 (if else)¶
newlist = [x if x != "banana" else "orange" for x in fruits]
if 和 for 的位置不是固定的，当 if else 成对出现时，放在 for 前面； 否则 if 放在后面（只有过滤作用）
二维 comprehension¶
用 comprehension 计算 9x9 乘法表中的所有计算结果，共 81 个乘法
for in 写法¶
for i in a:
for j in b:
c.append(i*j)
2d comprehension¶
a = range(1, 10)
b = range(1, 10)
c = [i * j for i in a for j in b]
print(c)
len(c)
要点：两个 for 是顺序写法
带 if 条件(三角形)¶
c = [i*j for i in a for j in b if i <= j]