Question

Alabama 1 Byrne，Bradley 68.16 68.16 0.0 LeFlore，Burton R. 31.71 31.71 0.0 未知0.13 0.13 0.0

我有一个类似的数据集：

STATE | DISTRICT | CANDIDATE NAME | GENERAL VOTE
Alabama | 1 | Byrne, Bradley |  68.16
Alabama | 1 | LeFlore, Burton R. | 31.71
Alabama | 1 | Unknown | 0.13
Alabama | 2 | Name | 65.43
Alabama | 2 | Name | 0.13

我必须按州和地区分组，因为每个州都有多个地区，而且有很多州。我已经这样做了。但是，我需要找到每个分组的最大值，并显示与此最大值一致的候选名称。我还必须显示每个分组中最大和最小一般投票之间的差异。我已经做了一些但是我被困了

 df_out = dfworking.groupby(["STATE", "D", "CANDIDATE NAME"])['GENERAL PERCENT'].agg(['max','min'])
df_out['Margin'] = df_out['max']-df_out['min']
df_new_out = dfworking.groupby(['STATE','D'])['GENERAL PERCENT'].max()

我不确定如何显示边距列，以及与同一数据框中的最大投票一致的名称。谢谢！

Answer 1

注意 - 必须先对STATE，DISTRICT和GENERAL VOTE列中的值进行排序。

#sorting
dfworking = dfworking.sort_values(['STATE','DISTRICT','GENERAL VOTE'], 
                                  ascending=[True, True, False])

#get index of max value in GENERAL VOTE column
df1 = dfworking.groupby(["STATE", "DISTRICT"])['GENERAL VOTE'].idxmax()
#create new column - not matched value return NaN
dfworking['cand'] = dfworking.loc[df1, 'CANDIDATE NAME']
#replace NaN by forward  filling
dfworking['cand'] = dfworking['cand'].ffill()

print (dfworking)
     STATE  DISTRICT      CANDIDATE NAME  GENERAL VOTE            cand
0  Alabama         1      Byrne, Bradley         68.16  Byrne, Bradley
1  Alabama         1  LeFlore, Burton R.         31.71  Byrne, Bradley
2  Alabama         1             Unknown          0.13  Byrne, Bradley
3  Alabama         2                Name         65.43            Name
4  Alabama         2                Name          0.13            Name

另一个解决方案是创建df与最高候选人join与原始人：

df1 = dfworking.loc[dfworking.groupby(["STATE", "DISTRICT"])['GENERAL VOTE'].idxmax()]
df1 = df1.set_index(['STATE','DISTRICT'])['CANDIDATE NAME'].rename('cand')
dfworking = dfworking.join(df1, on=['STATE','DISTRICT'])
print (dfworking)
     STATE  DISTRICT      CANDIDATE NAME  GENERAL VOTE            cand
0  Alabama         1      Byrne, Bradley         68.16  Byrne, Bradley
1  Alabama         1  LeFlore, Burton R.         31.71  Byrne, Bradley
2  Alabama         1             Unknown          0.13  Byrne, Bradley
3  Alabama         2                Name         65.43            Name
4  Alabama         2                Name          0.13            Name

Pandas在一列中找到最大值并从另一行显示

1 个答案: