分数是浮云，小结一下吧

ImN1 2013-06-23 12:44:23

备注：下面的方法只有少量是自己写的，大部分均为参考别人所得，由于只是平常笔记，没有一一记下原文出处了

Python中对列表list求交集

方法1
遍历b1,如果某个元素同时也存在于b2中,则返回
b1=[1,2,3]
b2=[2,3,4]
b3 = [val for val in b1 if val in b2]
print b3
运行结果如下
[2, 3]

方法2
把列表转换为集合,利用集合操作符求出交集,然后再转换回列表类型
b1=[1,2,3]
b2=[2,3,4]
b3=list(set(b1) & set(b2))
print b3
运行结果如下
[2, 3]
# 交为 & ，并为 | ，差为 -，不能使用 or

前面的例子中两个list都是简单的单元素列表,还有一种比较特殊的情况,就是有嵌套类型的
b1=[1,2,3]
b2=[[2,4],[3,5]]
b3 = [filter(lambda x: x in b1,sublist) for sublist in b2]
print b3
运行结果如下
[[2], [3]]

下面是文档中对filter的解释

filter(function, iterable)
Construct a list from those elements of iterable for which function returns true. iterable may be either a sequence, a container which supports iteration, or an iterator. If iterable is a string or a tuple, the result also has that type; otherwise it is always a list. If function is None, the identity function is assumed, that is, all elements of iterable that are false are removed.

Note that filter(function, iterable) is equivalent to [item for item in iterable if function(item)] if function is not None and [item for item in iterable if item] if function is None.

交换矩阵数组的行列
>>> a=[[1,2,3],[4,5,6]]
>>> list(zip(*a))
[(1, 4), (2, 5), (3, 6)]

求二维嵌套序列交集（部分单元相交）
a = [[8, 6, 3], [3, 4, 5], [5, 2, 7]]
b = [[2, 5, 4], [6, 8, 5]]
c = ((x, y) for x, y, z in b)
print([i for i in a if (i[1], i[0]) in c])

[[8, 6, 3], [5, 2, 7]]

字典求并集（覆盖，不覆盖则为x, y交换位置）
>>> x={'a':1, 'b':2}
>>> y={'a':4, 'c':3}
>>> dict(x, **y)
{'a': 4, 'b': 2, 'c': 3}

字典依据键名求交集
>>>dict((k,v) for k,v in x.items() if k in y)
{'a': 1}

字典依据键名求差集
>>>dict((k,v) for k,v in x.items() if k not in y)
{'b': 2}

字典行列交换
>>> a={'c': [11, 22, 33, 44], 'a': [1, 2, 3, 4], 'b': [5, 6, 7, 8]}
>>> [dict(zip(a, x)) for x in zip(*[a[y] for y in a])]
[{'a': 1, 'b': 5, 'c': 11}, {'a': 2, 'b': 6, 'c': 22}, {'a': 3, 'b': 7, 'c': 33}, {'a': 4, 'b': 8, 'c': 44}]

>>> k=tuple(a.keys())
>>> [dict(zip(k, x)) for x in zip(*[a[y] for y in k])]
[{'a': 1, 'b': 5, 'c': 11}, {'a': 2, 'b': 6, 'c': 22}, {'a': 3, 'b': 7, 'c': 33}, {'a': 4, 'b': 8, 'c': 44}]

In [167]: a={'a':[1,2,3,4], 'b':[5,6,7,8]}
In [168]: b = [{k: a[k][i] for k in a} for i in range(len(list(a.values())[0]))]
In [169]: b
Out[169]: [{'a': 1, 'b': 5}, {'a': 2, 'b': 6}, {'a': 3, 'b': 7}, {'a': 4, 'b': 8}]

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