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PYTHON中对列表list交集

luyued 发布于 2011-04-20 15:43   浏览 N 次  

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

python">dwapp@pttest1:/home/dwapp>python t1.py
[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

python">dwapp@pttest1:/home/dwapp>python t1.py
[2, 3]

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

python">dwapp@pttest1:/home/dwapp>python t1.py
[[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.

本文来自CSDN博客,转载请标明出处:http://blog.csdn.net/wh62592855/archive/2011/01/16/6145257.aspx

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