Question

我想做类似下面的事情。我想在列表推导中输出条件表达式。是否可以使用列表理解？

def why_bad(myvalue): #returns a list of reasons or an empty list it is good
   ...
   return [ reason1, reason2 ..]

bad_values = [ (myvalue,reasons) for myvalue in all_values if (reasons = why_bad(myvalue)) ]

Answer 1

你可以像这样创建你的列表理解，它返回值及其原因（或一个空列表），说明它为什么不好：

def why_bad(value):
    reasons = []
    if value % 2:
        reasons.append('not divisible by two')
    return reasons

all_values = [1,2,3]

bad_values = [(i, why_bad(i)) for i in all_values]
print bad_values

要扩展示例，您可以为每个不同的条件检查添加elif，以查找值为坏的原因并将其添加到列表中。

返回值：

[(1, ['not divisible by two']), (2, []), (3, ['not divisible by two'])]

如果all_values只有唯一值，您可以考虑创建字典而不是列表理解：

>>> bad_values = dict([(i, why_bad(i)) for i in all_values])
>>> print bad_values
{1: ['not divisible by two'], 2: [], 3: ['not divisible by two']}

Answer 2

您可以使用嵌套列表解析：

bad_values = [value_tuple for value_tuple in 
                  [(myvalue, why_bad(myvalue)) for myvalue in all_values]
              if value_tuple[1]] # value_tuple[1] == why_bad(myvalue)

或使用filter：

bad_values = filter(lambda value_tuple: value_tuple[1],
                    [(myvalue, why_bad(myvalue)) for myvalue in all_values])

Answer 3

不漂亮，但您可以尝试嵌套列表推导：

bad_values = [(v, reasons) for v in all_values 
                           for reasons in [why_bad(v)] if reasons]

python list_comprehension获取多个值

3 个答案: