获取groupby
元素的随机样本的最佳方法是什么?据我了解,groupby
只是一个可迭代的群组。
如果我想选择N = 200
元素,我将为迭代执行此操作的标准方法是:
rand = random.sample(data, N)
如果你尝试上面的数据是一个'分组'由于某种原因,结果列表的元素是元组。
我找到了以下示例,用于随机选择单个键groupby
的元素,但这不适用于多键groupby
。来自,How to access pandas groupby dataframe by key
创建groupby对象
grouped = df.groupby('some_key')
选择N个数据帧并获取其索引
sampled_df_i = random.sample(grouped.indices, N)
使用groupby对象抓取群组' get_group'方法
df_list = map(lambda df_i: grouped.get_group(df_i),sampled_df_i)
可选 - 将其全部转换回单个数据框对象
sampled_df = pd.concat(df_list, axis=0, join='outer')
答案 0 :(得分:8)
You can take a randoms sample of the unique values of df.some_key.unique()
, use that to slice the df
and finally groupby
on the resultant:
In [337]:
df = pd.DataFrame({'some_key': [0,1,2,3,0,1,2,3,0,1,2,3],
'val': [1,2,3,4,1,5,1,5,1,6,7,8]})
In [338]:
print df[df.some_key.isin(random.sample(df.some_key.unique(),2))].groupby('some_key').mean()
val
some_key
0 1.000000
2 3.666667
If there are more than one groupby keys:
In [358]:
df = pd.DataFrame({'some_key1':[0,1,2,3,0,1,2,3,0,1,2,3],
'some_key2':[0,0,0,0,1,1,1,1,2,2,2,2],
'val': [1,2,3,4,1,5,1,5,1,6,7,8]})
In [359]:
gby = df.groupby(['some_key1', 'some_key2'])
In [360]:
print gby.mean().ix[random.sample(gby.indices.keys(),2)]
val
some_key1 some_key2
1 1 5
3 2 8
But if you are just going to get the values of each group, you don't even need to groubpy
, MultiIndex
will do:
In [372]:
idx = random.sample(set(pd.MultiIndex.from_product((df.some_key1, df.some_key2)).tolist()),
2)
print df.set_index(['some_key1', 'some_key2']).ix[idx]
val
some_key1 some_key2
2 0 3
3 1 5
答案 1 :(得分:0)
我觉得较低级别的 @Entity()
@Unique(['a', 'b'])
export class AB {
@PrimaryGeneratedColumn('uuid')
uuid: string;
@ManyToOne(() => A, { nullable: false })
a: A;
@ManyToOne(() => B, { nullable: false })
b: B;
}
操作更简洁:
numpy