将For循环转换为List Comprehension：测试列表2中的元素是否与列表1中的元素部分匹配 [英] Convert For Loop to List Comprehension: Testing if elements in list 2 are partial match for elements in list 1

问题描述

有些python /编程新手在这里，

我已经想出了一个2嵌套for循环代码来测试第二循环的元素是第一个列表中元素的部分（或完整）匹配，如果是的话，这些元素将从第一个列表中删除。这里是代码：

 >>> lst = [我的名字是Bob Jenkins，Tommy Cooper I am，I be Bazil，I Tarzan] 
>>> names = [Bob，Tarzan] 
>>> for lst lst：
 for n in name：
 if n in l：
 lst.remove（l）
 
 
>>> ; print lst 
 ['Tommy Cooper I'，'I be Bazil'] 
  

这里的问题是，我并不想要修改 lst ，但创建一个新的列表。因此，列表理解提高了头。但是，我还没有弄清楚列表理解是否能解决这个问题。我已经尝试了 [我为我在lst中如果n不在我名字中n] 但是这不起作用。当然， .remove 方法不适用于列表理解。过去一个小时我一直困惑，却不能拿出任何东西。

有什么想法？

解决方案

/ p>

  lst = [我的名字是Bob Jenkins，Tommy Cooper I am，我是Bazil，I Tarzan ] 
 names = [Bob，Tarzan] 
 
 new = [a如果全部（x不在x中名称中）] 
打印新
  

somewhat of a python/programming newbie here,

I have come up with a code of 2 nested for loops to test if the elements of the the 2nd loop are a partial (or complete) match of the elements in the first list, and if so those elements are removed from the 1st list. Here is the code:

>>> lst = ["my name is Bob Jenkins", "Tommy Cooper I am", "I be Bazil", "I Tarzan"]
>>> names = ["Bob", "Tarzan"]
>>> for l in lst:
        for n in names:
            if n in l:
                lst.remove(l)


>>> print lst
['Tommy Cooper I am', 'I be Bazil']

The problem here is that I don't actual want to modify lst, but create a new list. Thus list comprehension raises its head. However, I have not been able to figure out a list comprehension to pull this off. I have tried [i for i in lst for n in names if n not in i] but this does not work. Of course the .remove method is not for list comprehension. I have been puzzling for the past hour, but cannot come up with anything.

Any ideas?

解决方案

There you go :) :

lst = ["my name is Bob Jenkins", "Tommy Cooper I am", "I be Bazil", "I Tarzan"]
names = ["Bob", "Tarzan"]

new=[a for a in lst if all(x not in a for x in names) ]
print new

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