Question

我有

iList = [('FirstParam', 1), ('FirstParam', 2), 
         ('FirstParam', 3), ('FirstParam', 4), 
         ('SecondParam', 5), ('SecondParam', 6), 
         ('SecondParam', 7)]

我正在努力形成：

iList1 = [('FirstParam', 1), ('FirstParam', 2), 
         ('FirstParam', 3), ('FirstParam', 4)]

iList2 = [('SecondParam', 5), ('SecondParam', 6), 
         ('SecondParam', 7)]

我试过了：

for key, group in itertools.groupby(iList, operator.itemgetter(0)):
    print "List",(list(group))

它以我想要的方式打印2个列表。输出：

List [('FirstParam', 1), ('FirstParam', 2), 
         ('FirstParam', 3), ('FirstParam', 4)]
List [('SecondParam', 5), ('SecondParam', 6), 
         ('SecondParam', 7)]

我担心的是我应该如何制作iList1和iList2？我没有按照(list(group))的方式工作并打印列表。

Answer 1

您可以使用列表理解并解压缩到iList1和iList2：

iList1, iList2 = [list(group) for key, group in itertools.groupby(iList, operator.itemgetter(0))]

然后你会得到你想要的东西：

>>> iList1
[('FirstParam', 1), ('FirstParam', 2), ('FirstParam', 3), ('FirstParam', 4)]
>>> iList2
[('SecondParam', 5), ('SecondParam', 6), ('SecondParam', 7)]

我应该如何打破元组列表，以便它形成具有相同键的列表？

1 个答案: