![](/img/trans.png)
[英]How to remove all elements matching a specific criteria from a list of lists?
[英]Remove all matching value from a list of lists
我有以下列表对象列表
myList = [[123,0.0,345,0.0,0.0,0.0],
[45,0.0,0.0,0.0],
[67,8,0.0,5,6,7,0.0]
我想从此列表中删除所有零。
我遵循了这个问题,并按如下所示进行编码。
myList = list(filter(lambda j:j!=0,myList[i]) for i in range(len(myList)))
但是我正在获取过滤器对象列表作为输出。 代码中有什么错误。
[<filter object at 0x7fe7bdfff8d0>, <filter object at 0x7fe7a6eaaf98>, <filter object at 0x7fe7a6f08048>,
您忘记使用list
强制转换内部filter
功能,执行此操作时,代码将按预期工作:)
myList = [[123,0.0,345,0.0,0.0,0.0],
[45,0.0,0.0,0.0],
[67,8,0.0,5,6,7,0.0]]
#Cast inner filter into a list
myList = list(list(filter(lambda j:j!=0,myList[i])) for i in range(len(myList)))
print(myList)
输出将是
[[123, 345], [45], [67, 8, 5, 6, 7]]
同样,更简单的理解方式是使用列表理解
myList = [[123,0.0,345,0.0,0.0,0.0],
[45,0.0,0.0,0.0],
[67,8,0.0,5,6,7,0.0]]
#Using list comprehension, in the inner loop check if item is non-zero
myList = [ [item for item in li if item != 0] for li in myList ]
print(myList)
输出将是
[[123, 345], [45], [67, 8, 5, 6, 7]]
尝试这个 :
newList = [list(filter(lambda j:j!=0, i)) for i in myList]
输出 :
[[123, 345], [45], [67, 8, 5, 6, 7]]
您还可以通过列表理解来做到这一点:
cleaned = [ [e for e in row if e != 0] for row in myList ]
您只需要包装过滤器,而不是整个语句:
myList = [list(filter(lambda j:j!=0,myList[i]) for i in range(len(myList))]
另外,您可以跳过索引,并通过myList
的列表进行myList
:
myList = [list(filter(lambda j:j!=0, inner_list) for inner_list in myList]
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.