[英]Creating a dictionary with multiple values per key from a list of dictionaries
I have the following list of dictionaries: 我有以下词典列表:
listofdics = [{'StrId': 11, 'ProjId': 1},{'StrId': 11,'ProjId': 2},
{'StrId': 22, 'ProjId': 3},{'StrId': 22, 'ProjId': 4},
{'StrId': 33, 'ProjId': 5},{'StrId': 33, 'ProjId': 6},
{'StrId': 34, 'ProjId': 7}]
I need to get all ProjId
values for StrId
that are duplicate. 我需要把所有
ProjId
的值StrId
是重复的。 So this is the output I'm looking for: 所以这是我正在寻找的输出:
new_listofdics = [{11:[1,2]}, {22:[3,4]}, {33:[5,6]], {34:[7]}]
I wrote a function that creates a list of dictionaries with StrId
values as keys, and a list with all ProjId
that share the same key as values. 我编写了一个函数,该函数创建一个将
StrId
值作为键的字典列表,以及一个将所有ProjId
与值共享相同键的列表。 Here it is: 这里是:
def compare_projids(listofdics):
proj_ids_dups = {}
for row in listofdics:
id_value = row['StrId']
proj_id_value = row['ProjId']
proj_ids_dups[id_value]=proj_id_value
if row['StrId'] == id_value:
sum_ids = []
sum_ids.append(proj_id_value)
proj_ids_dups[id_value]=sum_ids
return proj_ids_dups
This is the output I get now: 这是我现在得到的输出:
new_listofdics= {33: [6], 34: [7], 11: [2], 22: [4]}
What I see is that append
replaces each ProjId
value with the last one iterated, instead of adding them at the end of the list. 我看到的是,
append
用迭代的最后一个替换每个ProjId
值,而不是将它们添加到列表的末尾。
How can I fix this?... 我怎样才能解决这个问题?...
It's unclear why you need to have such output new_listofdics = [{11:[1,2]}, {22:[3,4]}, {33:[5,6]], {34:[7]}]
, because it's better to have just dict
object. 目前尚不清楚为什么需要这样的输出
new_listofdics = [{11:[1,2]}, {22:[3,4]}, {33:[5,6]], {34:[7]}]
,因为最好只有dict
对象。
So program would look like this 所以程序看起来像这样
>>> from collections import defaultdict
>>> listofdics = [{'StrId': 11, 'ProjId': 1},{'StrId': 11,'ProjId': 2},
{'StrId': 22, 'ProjId': 3},{'StrId': 22, 'ProjId': 4},
{'StrId': 33, 'ProjId': 5},{'StrId': 33, 'ProjId': 6},
{'StrId': 34, 'ProjId': 7}]
>>> output = defaultdict(list)
>>> for item in listofdics:
... output[item.get('StrId')].append(item.get('ProjId'))
>>> dict(output)
{11: [1, 2], 22: [3, 4], 33: [5, 6], 34: [7]}
It's much easier to go through that dict that your desired output. 通过该命令获得所需的输出要容易得多。
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