找到每行具有最大值的列索引

Question

我有以下数据框：

   Name1 Scr1 Name2 Scr2 Name3 Scr3
   NY    21   CA    45   SF    37
   AZ    31   BK    46   AK    23

我正在尝试获取每行的最大值和每行的相应名称：

df.idxmax(axis=1)

但是如何获得相应的名称呢？

预期输出：

   Name Hi_Scr
   CA    45
   BK    46

Answer 1

使用：

通过 filter, convert values to numpy array by values

Scr

使用 argmax
通过索引

Name

获取数值的最大值
通过构造函数创建DataFrame

a = df.filter(like='Scr').values
b = a.argmax(axis=1)
c = df.filter(like='Name').values[np.arange(len(df.index)), b]
d = a.max(axis=1)

df = pd.DataFrame({'Name':c, 'Hi_Scr':d}, columns=['Name','Hi_Scr'])
print (df)
  Name  Hi_Scr
0   CA      45
1   BK      46

Pandas 解决方案非常相似 - 通过 extract, then select by xs and for looking values use lookup:

在列中创建 MultiIndex

a = df.columns.to_series().str.extract('(\D+)(\d+)', expand=False)
df.columns = pd.MultiIndex.from_tuples(a.values.tolist())

a = df.xs('Scr', axis=1)
b = a.idxmax(axis=1)
c = df.xs('Name', axis=1).lookup(df.index, b)
d = a.max(axis=1)

df = pd.DataFrame({'Name':c, 'Hi_Scr':d}, columns=['Name','Hi_Scr'])
print (df)
  Name  Hi_Scr
0   CA      45
1   BK      46

时间:

df = pd.concat([df]*10000).reset_index(drop=True)


def jez2(df):
    a = df.columns.to_series().str.extract('(\D+)(\d+)', expand=False)
    df.columns = pd.MultiIndex.from_tuples(a.values.tolist())

    a = df.xs('Scr', axis=1)
    b = a.idxmax(axis=1)
    c = df.xs('Name', axis=1).lookup(df.index, b)
    d = a.max(axis=1)

    return pd.DataFrame({'Name':c, 'Hi_Scr':d}, columns=['Name','Hi_Scr'])


def jez1(df):
    a = df.filter(like='Scr').values
    b = a.argmax(axis=1)
    c = df.filter(like='Name').values[np.arange(len(df.index)), b]
    d = a.max(axis=1)

    return  pd.DataFrame({'Name':c, 'Hi_Scr':d}, columns=['Name','Hi_Scr'])

def dark(df):
    df['id'] = df.index
    ndf = pd.wide_to_long(df, ["Name", "Scr"], i="id", j="number").reset_index(0).set_index('Name')
    return ndf.groupby('id')['Scr'].agg(['max','idxmax']).rename(columns= {'max':'Hi_Scr','idxmax':'Name'})

import time

t0 = time.time()
print (jez1(df).head())
t1 = time.time() - t0
print (t1)
print (dark(df).head())
t2 = time.time() - t1
print (t2)
print (jez2(df).head())
t3 = time.time() - t2
print (t3)

  Name  Hi_Scr
0   CA      45
1   BK      46
2   CA      45
3   BK      46
4   CA      45
#jez1 solution
0.015599966049194336
    Hi_Scr Name
id             
0       45   CA
1       46   BK
2       45   CA
3       46   BK
4       45   CA
#dark solution
1515070100.961423
  Name  Hi_Scr
0   CA      45
1   BK      46
2   CA      45
3   BK      46
4   CA      45
#jez2 solution
0.04679989814758301

Answer 2

我会用 pd.wide_to_long 这样做：

df['id'] = df.index
ndf = pd.wide_to_long(df, ["Name", "Scr"], i="id", j="number").reset_index(0).set_index('Name')

#       id  Scr
# Name         
# NY     0   21
# AZ     1   31
# CA     0   45
# BK     1   46
# SF     0   37
# AK     1   23

# Thank you @jezrael for the improvement
ndf.groupby('id')['Scr'].agg(['max','idxmax']).rename(columns= {'max':'Hi_Scr','idxmax':'Name'})

   Name  Hi Scr
id             
0    CA      45
1    BK      46

Answer 3

类似于

df1=df.select_dtypes(include=[object])
df2=df.select_dtypes(exclude=[object])
pd.DataFrame({'Name':df1.values[np.where(df2.eq(df2.max(1),0))],'Scr':df2.max(1)})

Out[342]: 
  Name  Scr
0   CA   45
1   BK   46

找到每行具有最大值的列索引

Find the column index which has the maximum value for each row

pandas

argmax