Question

我从数据库返回的记录如下所示：

region       month_taken         total_att num_classes
Colorado    2013-01-01 00:00:00.000 78485   4648
Colorado    2013-02-01 00:00:00.000 71769   4162
Midwest     2013-01-01 00:00:00.000 110508  7101
Midwest     2013-02-01 00:00:00.000 103545  6410

我试图让他们进入列表：

Total_att：

[{"data": [78485, 71769], "name": "Colorado"}, {"data": [110508, 103545], "name": "Midwest"}]

num_classes：

[{"data": [4648, 4162], "name": "Colorado"}, {"data": [7101, 6410], "name": "Midwest"}]

我发现了itertools.groupby，它可以完成我想要的工作，但是我在使用多个值列表时遇到困难（因为没有更好的术语）。

totalResults = []            
for key, location in groupby(rows, lambda k: k[0]):
    totalRow = dict()
    totalRow['name'] = key
    totalRow['data'] = [x[2] for x in location]
    totalResults.append(totalRow)

太棒了，这让我得到了我的total_att列表，但后来又做了一个额外的groupby循环来创建＆＃34; num_classes＆＃34;列表，这看起来很荒谬。我在文档中看到了这一点，但老实说，如果我将其转换为列表，我不太确定它的含义或如何处理我的问题：

The returned group is itself an iterator that shares the underlying iterable with groupby(). Because the source is shared, when the groupby() object is advanced, the previous group is no longer visible. So, if that data is needed later, it should be stored as a list:

那么，我如何创建我的列表而不用为groupby中的键，位置（行，lambda k：k [0]）做多个：？

我希望这很清楚，但我很乐意在必要时提供更多信息。

Answer 1

totalResults = [] 
totalClasses = []           
for key, location in groupby(rows, lambda k: k[0]):
    location = list(location)
    totalResults.append(dict(name=key, data=[x[2] for x in location]))
    totalClasses.append(dict(name=key, data=[x[3] for x in location]))

Python itertools.groupby多个值

1 个答案: