我应该如何打破元组列表，使其形成具有相同键的列表？

Question

我有

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

我试图形成：

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

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

我试过：

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

它打印出我想要的两张单子。 输出：

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

我关心的是，我应该如何使它们成为iList1和iList2？ 我不懂(list(group))工作和打印列表的方式

Answer 1

您可以使用列表理解并将其分解为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：

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