获取 Pandas 数据框中每一列的唯一值 - 帮助我创建更小的更易于管理的数据框来执行指标
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getting the unique values of every column in a pandas dataframe - to help me create smaller more manageable dataframes to perform metrics on
提问by yoshiserry
I started off wanting to turn a column from a pandas dataframe into a list, and then get the unique values, with the aim of iterating over those unique values in a for loop, and creating a few smaller dataframes. I.e. one for each cluster. Then I want to store these smaller dataframes in a dictionary object.
我开始想将 Pandas 数据帧中的一列转换为列表，然后获取唯一值，目的是在 for 循环中迭代这些唯一值，并创建一些较小的数据帧。即每个集群一个。然后我想将这些较小的数据帧存储在字典对象中。
@ben suggested I start a new question and ask about the GroupBy Method of pandas dataframes to perform this task?
@ben 建议我开始一个新问题并询问 Pandas 数据帧的 GroupBy 方法来执行此任务？
My original post is over here: get list from pandas dataframe column
我的原始帖子在这里： 从Pandas数据框列中获取列表
My Data:
cluster load_date budget actual fixed_price
A 1/1/2014 1000 4000 Y
A 2/1/2014 12000 10000 Y
A 3/1/2014 36000 2000 Y
B 4/1/2014 15000 10000 N
B 4/1/2014 12000 11500 N
B 4/1/2014 90000 11000 N
C 7/1/2014 22000 18000 N
C 8/1/2014 30000 28960 N
C 9/1/2014 53000 51200 N
For example: for item in cluster_list(where cluster list is the unique set of values in cluster)
例如：对于 cluster_list 中的项目（其中集群列表是集群中唯一的一组值）
create a dataframe for cluster a, where budget > X etc
Then do the same for the other clusters, and put them in a dictionary.
然后对其他集群做同样的事情，并将它们放入字典中。
Then be able to get a certain dataframe out of the dictionary, say only the dataframe for cluster B where budget > X
然后能够从字典中获取某个数据帧，仅说预算 > X 的集群 B 的数据帧
GetDf(key):
return dict(key)
Thanks in advance
提前致谢
回答by Andy Hayden
There's two parts to this question. First, filter those columns where budget < X:
这个问题有两个部分。首先，过滤那些预算 < X 的列：
In [11]: df1 = df[df['budget'] > 10000]
In [12]: df1
Out[12]:
cluster load_date budget actual fixed_price
1 A 2/1/2014 12000 10000 Y
2 A 3/1/2014 36000 2000 Y
3 B 4/1/2014 15000 10000 N
4 B 4/1/2014 12000 11500 N
5 B 4/1/2014 90000 11000 N
6 C 7/1/2014 22000 18000 N
7 C 8/1/2014 30000 28960 N
8 C 9/1/2014 53000 51200 N
Now you can groupby cluster, and get the groups:
现在您可以按集群分组，并获取组：
In [13]: g = df1.groupby('cluster')
In [14]: g.get_group('A')
Out[14]:
cluster load_date budget actual fixed_price
1 A 2/1/2014 12000 10000 Y
2 A 3/1/2014 36000 2000 Y
Note: if you reallywant a dictionary then you can use:
注意：如果你真的想要一本字典，那么你可以使用：
In [15]: d = dict(iter(g))
In [16]: d['A']
Out[16]:
cluster load_date budget actual fixed_price
1 A 2/1/2014 12000 10000 Y
2 A 3/1/2014 36000 2000 Y
