如何在巨大数据框的每一行中查找前n个值的列索引 [英] How to find column-index of top-n values within each row of huge dataframe

问题描述

我的数据框的格式为:(示例数据)

I have a dataframe of format: (example data)

      Metric1  Metric2  Metric3  Metric4  Metric5
ID    
1     0.5      0.3      0.2      0.8      0.7    
2     0.1      0.8      0.5      0.2      0.4    
3     0.3      0.1      0.7      0.4      0.2    
4     0.9      0.4      0.8      0.5      0.2    

其中分数范围在[0,1]和我希望生成的函数之间，对于每个id(行)，该函数计算前n个指标，其中n是函数以及原始数据帧的输入.

where score range between [0,1] and I wish to generate a function that, for each id (row), calculates the top n metrics, where n is an input of the function along with the original dataframe.

我的理想输出是:(例如n = 3)

My ideal output would be:(for eg. n = 3)

      Top_1     Top_2     Top_3
ID    
1     Metric4   Metric5   Metric1    
2     Metric2   Metric3   Metric5    
3     Metric3   Metric4   Metric1    
4     Metric1   Metric3   Metric4  

现在，我编写了一个可以正常工作的函数:

Now I have written a function that does work:

def top_n_partners(scores,top_n=3):
metrics = np.array(scores.columns)
records=[]
for rec in scores.to_records():
    rec = list(rec)
    ID = rec[0]
    score_vals = rec[1:]
    inds = np.argsort(score_vals)
    top_metrics = metrics[inds][::-1]
    dic = {
        'top_score_%s' % (i+1):top_metrics[i]
        for i in range(top_n)
    }
    dic['ID'] = ID
    records.append(dic)
top_n_df = pd.DataFrame(records)
top_n_df.set_index('ID',inplace=True)
return top_n_df

但是，对于我要运行的数据量(具有数百万行的数据帧)而言，它似乎效率很低/很慢，我想知道是否有更聪明的方法来解决这个问题?

However it seems rather inefficient/slow especially for the volume of data I'd be running this over (dataframe with millions of rows) and I was wondering if there was a smarter way to go about this?

推荐答案

您可以使用 numpy.argsort :

You can use numpy.argsort:

print (np.argsort(-df.values, axis=1)[:,:3])
[[3 4 0]
 [1 2 4]
 [2 3 0]
 [0 2 3]]

print (df.columns[np.argsort(-df.values, axis=1)[:,:3]])

Index([['Metric4', 'Metric5', 'Metric1'], ['Metric2', 'Metric3', 'Metric5'],
       ['Metric3', 'Metric4', 'Metric1'], ['Metric1', 'Metric3', 'Metric4']],
      dtype='object')

df = pd.DataFrame(df.columns[np.argsort(-df.values, axis=1)[:,:3]], 
                               index=df.index)
df = df.rename(columns = lambda x: 'Top_{}'.format(x + 1))
print (df)
      Top_1    Top_2    Top_3
ID                           
1   Metric4  Metric5  Metric1
2   Metric2  Metric3  Metric5
3   Metric3  Metric4  Metric1
4   Metric1  Metric3  Metric4 

谢谢您 Divakar 用于改进:

n = 3
df = pd.DataFrame(df.columns[df.values.argsort(1)[:,-n+2:1:-1]], 
                               index=df.index)

df = df.rename(columns = lambda x: 'Top_{}'.format(x + 1))
print (df)
      Top_1    Top_2    Top_3
ID                           
1   Metric4  Metric5  Metric1
2   Metric2  Metric3  Metric5
3   Metric3  Metric4  Metric1
4   Metric1  Metric3  Metric4                

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