[英]Iterate over list of tuples and unpack first elements only
Assume the following list: foo = [(1, 2, 3, 4), (5, 6, 7, 8)]
假设有以下列表:
foo = [(1, 2, 3, 4), (5, 6, 7, 8)]
Is there a way to iterate over the list and unpack the first two elements of the inner tuple only?有没有办法遍历列表并仅解包内部元组的前两个元素?
This is a usual pattern: {a: b for a, b, _, _ in foo}
, but this breaks if foo
is modified (program change) and the tuple now contains 5 elements instead of 4 (the the list comprehension would need to be modified accordingly).这是一个通常的模式:
{a: b for a, b, _, _ in foo}
,但是如果foo
被修改(程序更改)并且元组现在包含 5 个元素而不是 4 个(列表理解需要作相应修改)。 I really like to name the elements instead of calling {f[0]: f[1] for f in foo}
, so ideally, there would be some sort of "absorb all not unpacked variable", so one could call {a: b for a, b, absorb_rest in foo}
.我真的很喜欢命名元素而不是调用
{f[0]: f[1] for f in foo}
,所以理想情况下,会有某种“吸收所有未解压的变量”,所以可以调用{a: b for a, b, absorb_rest in foo}
。 If that's possible, it wouldn't matter how many elements are contained in the tuple (as long as there are at least 2).如果可能的话,元组中包含多少个元素并不重要(只要至少有 2 个)。
You can use extended iterable unpacking , where you extract the first two elements of the tuple, and ignore the rest of the elements.您可以使用扩展的可迭代解包,在其中提取元组的前两个元素,并忽略其余元素。 Note that this only works for
python3
请注意,这仅适用于
python3
{a:b for a, b, *c in foo}
You can use extended iterable unpacking , keeping in this way the first two values from the iterable and ignoring the rest:您可以使用扩展的可迭代解包,以这种方式保留可迭代的前两个值并忽略其余值:
{a: b for a, b, *_ in foo}
# {1: 2, 5: 6}
Try this :尝试这个 :
dict_ = {a:b for a,b, *_ in foo}
Output :输出:
{1: 2, 5: 6}
If foo
is changed to [(1, 2, 3, 4,9,0), (5, 6, 7, 8, 11, 16)]
by program later, dict_
still remains : {1: 2, 5: 6}
如果
foo
稍后被程序更改为[(1, 2, 3, 4,9,0), (5, 6, 7, 8, 11, 16)]
, dict_
仍然保持: {1: 2, 5: 6}
use lambda
使用
lambda
foo = [(1, 2, 3, 4), (5, 6, 7, 8)]
output = list(map(lambda x:{x[0]:x[1]}, foo))
print(output)
output输出
[{1: 2}, {5: 6}]
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.