2017/09/05 09:08
identity 就是单位函数：lambda *args: args
2017/08/07 16:04
python3 中在显示map时要做list操作
list(map(abc, list1))
2017/01/19 16:52

2016/08/15 21:37
1. map(function, iterable)
=> [function(i) for i in iterable]
2. map(function, iterable, ...) => map(function, *iterable_list)
=> [function(*i) for i in iterable_list]

2016/08/15 21:37
1. map(function, iterable)
=> [function(i) for i in iterable]
2. map(function, iterable, ...) => map(function, *iterable_list)
=> [function(*i) for i in iterable_list]

2016/04/08 10:06

2015/12/30 17:17

return x+100
list1=[11,22,33]

2015/12/29 14:03

return x+100
list1=[11,22,33]
2015/07/21 10:45

#### 引用来自“songzhenya”的评论

[abc(a,b,c) for a,b,c in (list1,list2,list3)]
it's not right, look 5 floor
2015/07/21 10:45

#### 引用来自“songzhenya”的评论

[abc(a,b,c) for a,b,c in (list1,list2,list3)]
it's not right, look 5 floor
2014/12/04 11:38

2014/11/19 11:41
[abc(a[i],b[i],c[i]) for i in xrange(len(a))]
2014/11/10 15:05
[abc(a,b,c) for a in list1 for b in list2 for c in list3]这种写法是错误的，歪曲了map()函数的实现过程，看看源码就知道了
2014/11/10 11:52
[abc(a,b,c) for a,b,c in (list1,list2,list3)]
2014/04/17 13:28

[abc(a,b,c) for a,b,c in zip(list1,list2,list3)]
2014/03/04 14:24

2013/09/29 19:00
#这个例子我们在上面看过了，若是用列表推导应该怎么写呢？我想是这样的：
#[abc(a,b,c) for a in list1 for b in list2 for c in list3]

[abc(list1[i],list2[i],list3[i]) for i in xrange(3)]

2013/03/19 13:38

#### 引用来自“Yoker”的评论

2013/03/19 12:41

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