还有另一种(易于推广的)方法，其基础是pandas.DataFrame.idxmax。首先，易于概括的序言。

# Indeed, all your conditions boils down to the following
_gt_1_key = 'two_or_more'
_lt_1_key = 'other'

# The "dictionary-based" if-else statements
labels = {
    _gt_1_key     : 'Two Or More',
    'eri_hispanic': 'Hispanic',
    'eri_nat_amer': 'A/I AK Native',
    'eri_asian'   : 'Asian',
    'eri_afr_amer': 'Black/AA',
    'eri_hawaiian': 'Haw/Pac Isl.',
    'eri_white'   : 'White',  
    _lt_1_key     : 'Other',
}

# The output-driving 1-0 matrix
mat = df.filter(regex='^eri_').copy()  # `~.copy` to avoid `SettingWithCopyWarning`

．.． 最后，以向量化的方式:

mat[_gt_1_key] = gt1 = mat.sum(axis=1)
mat[_lt_1_key] = gt1.eq(0).astype(int)
race_label     = mat.idxmax(axis=1).map(labels)

在哪里

>>> race_label
0           White
1        Hispanic
2           White
3           White
4           Other
5           White
6     Two Or More
7           White
8    Haw/Pac Isl.
9           White
dtype: object

那是一只熊猫。您可以轻松地在df中托管系列实例，即df['race_label'] = race_label。

上面的答案是完全有效的，但是存在一个向量化的解决方案，形式为numpy.select。这允许你定义条件，然后定义这些条件的输出，比使用apply更有效:

---

首先，定义条件:

conditions = [
    df['eri_hispanic'] == 1,
    df[['eri_afr_amer', 'eri_asian', 'eri_hawaiian', 'eri_nat_amer', 'eri_white']].sum(1).gt(1),
    df['eri_nat_amer'] == 1,
    df['eri_asian'] == 1,
    df['eri_afr_amer'] == 1,
    df['eri_hawaiian'] == 1,
    df['eri_white'] == 1,
]

现在，定义相应的输出:

outputs = [
    'Hispanic', 'Two Or More', 'A/I AK Native', 'Asian', 'Black/AA', 'Haw/Pac Isl.', 'White'
]

最后，使用numpy.select:

res = np.select(conditions, outputs, 'Other')
pd.Series(res)

0           White
1        Hispanic
2           White
3           White
4           Other
5           White
6     Two Or More
7           White
8    Haw/Pac Isl.
9           White
dtype: object

---

为什么要numpy。选择使用而不是应用?下面是一些性能检查:

df = pd.concat([df]*1000)

In [42]: %timeit df.apply(lambda row: label_race(row), axis=1)
1.07 s ± 4.16 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

In [44]: %%timeit
    ...: conditions = [
    ...:     df['eri_hispanic'] == 1,
    ...:     df[['eri_afr_amer', 'eri_asian', 'eri_hawaiian', 'eri_nat_amer', 'eri_white']].sum(1).gt(1),
    ...:     df['eri_nat_amer'] == 1,
    ...:     df['eri_asian'] == 1,
    ...:     df['eri_afr_amer'] == 1,
    ...:     df['eri_hawaiian'] == 1,
    ...:     df['eri_white'] == 1,
    ...: ]
    ...:
    ...: outputs = [
    ...:     'Hispanic', 'Two Or More', 'A/I AK Native', 'Asian', 'Black/AA', 'Haw/Pac Isl.', 'White'
    ...: ]
    ...:
    ...: np.select(conditions, outputs, 'Other')
    ...:
    ...:
3.09 ms ± 17 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

使用numpy。Select极大地提高了性能，并且随着数据的增长，差异只会增加。

如果我们检查它的源代码，apply()是Python for循环的语法糖(通过FrameApply类的apply_series_generator()方法)。因为它有pandas开销，所以通常比Python循环要慢。

尽可能使用优化的(向量化的)方法。如果必须使用循环，请使用@numba。jit装饰。

1. 不要使用apply()作为if-else阶梯

Df.apply()是在pandas中最慢的方法。如user3483203和Mohamed Thasin ah的回答所示，根据数据帧大小，np.select()和df. select()。要产生相同的输出，Loc可能比df.apply()快50-300倍。

碰巧，使用numba模块的@jit装饰器的循环实现(与apply()类似)比df快(大约50-60%)。Loc和np.select.1

Numba在numpy数组上工作，因此在使用jit装饰器之前，需要将数据帧转换为numpy数组。然后通过检查循环中的条件在预先初始化的空数组中填充值。由于numpy数组没有列名，所以必须通过循环中的索引访问列。与apply()中的if-else天梯相比，jit函数中的if-else天梯最不方便的部分是通过索引访问列。否则，它几乎是相同的实现。

import numpy as np
import numba as nb
@nb.jit(nopython=True)
def conditional_assignment(arr, res):    
    length = len(arr)
    for i in range(length):
        if arr[i][3] == 1:
            res[i] = 'Hispanic'
        elif arr[i][0] + arr[i][1] + arr[i][2] + arr[i][4] + arr[i][5] > 1:
            res[i] = 'Two Or More'
        elif arr[i][0]  == 1:
            res[i] = 'Black/AA'
        elif arr[i][1] == 1:
            res[i] = 'Asian'
        elif arr[i][2] == 1:
            res[i] = 'Haw/Pac Isl.'
        elif arr[i][4] == 1:
            res[i] = 'A/I AK Native'
        elif arr[i][5] == 1:
            res[i] = 'White'
        else:
            res[i] = 'Other'
    return res

# the columns with the boolean data
cols = [c for c in df.columns if c.startswith('eri_')]
# initialize an empty array to be filled in a loop
# for string dtype arrays, we need to know the length of the longest string
# and use it to set the dtype
res = np.empty(len(df), dtype=f"<U{len('A/I AK Native')}")
# pass the underlying numpy array of `df[cols]` into the jitted function
df['rno_defined'] = conditional_assignment(df[cols].values, res)

---

2. 不要使用apply()进行数值操作

如果需要通过添加两列来添加新行，您的第一反应可能是写入

df['c'] = df.apply(lambda row: row['a'] + row['b'], axis=1)

但与此相反，使用sum(axis=1)方法逐行添加(如果只有两列，则使用+运算符):

df['c'] = df[['a','b']].sum(axis=1)
# equivalently
df['c'] = df['a'] + df['b']

根据数据帧的大小，sum(1)可能比apply()快100倍。

事实上，你几乎不需要apply()在pandas数据帧上进行数值操作，因为它已经优化了大多数操作的方法:加法(sum(1))、减法(sub()或diff())、乘法(prod(1))、除法(div()或/)、幂(pow())、>、>=、==、%、//、&、|等都可以在整个数据帧上执行，而不需要apply()。

例如，假设你想用下面的规则创建一个新列:

IF [colC] > 0 THEN RETURN [colA] * [colB]
ELSE RETURN [colA] / [colB]

使用优化的pandas方法，可以写成

df['new'] = df[['colA','colB']].prod(1).where(df['colC']>0, df['colA'] / df['colB'])

等效的apply()解决方案是:

df['new'] = df.apply(lambda row: row.colA * row.colB if row.colC > 0 else row.colA / row.colB, axis=1)

对于具有20k行的数据帧，使用优化方法的方法比等效的apply()方法快250倍。这种差距只会随着数据大小的增加而增加(对于具有1 mil行的数据帧，它要快365倍)，并且时间差将变得越来越明显

---

1

: In the below result, I show the performance of the three approaches using a dataframe with 24 mil rows (this is the largest frame I can construct on my machine). For smaller frames, the numba-jitted function consistently runs at least 50% faster than the other two as well (you can check yourself).

def pd_loc(df):
    df['rno_defined'] = 'Other'
    df.loc[df['eri_nat_amer'] == 1, 'rno_defined'] = 'A/I AK Native'
    df.loc[df['eri_asian'] == 1, 'rno_defined'] = 'Asian'
    df.loc[df['eri_afr_amer'] == 1, 'rno_defined'] = 'Black/AA'
    df.loc[df['eri_hawaiian'] == 1, 'rno_defined'] = 'Haw/Pac Isl.'
    df.loc[df['eri_white'] == 1, 'rno_defined'] = 'White'
    df.loc[df[['eri_afr_amer', 'eri_asian', 'eri_hawaiian', 'eri_nat_amer', 'eri_white']].sum(1) > 1, 'rno_defined'] = 'Two Or More'
    df.loc[df['eri_hispanic'] == 1, 'rno_defined'] = 'Hispanic'
    return df

def np_select(df):
    conditions = [df['eri_hispanic'] == 1,
                  df[['eri_afr_amer', 'eri_asian', 'eri_hawaiian', 'eri_nat_amer', 'eri_white']].sum(1).gt(1),
                  df['eri_nat_amer'] == 1,
                  df['eri_asian'] == 1,
                  df['eri_afr_amer'] == 1,
                  df['eri_hawaiian'] == 1,
                  df['eri_white'] == 1]
    outputs = ['Hispanic', 'Two Or More', 'A/I AK Native', 'Asian', 'Black/AA', 'Haw/Pac Isl.', 'White']
    df['rno_defined'] = np.select(conditions, outputs, 'Other')
    return df


@nb.jit(nopython=True)
def conditional_assignment(arr, res):
    
    length = len(arr)
    for i in range(length):
        if arr[i][3] == 1 :
            res[i] = 'Hispanic'
        elif arr[i][0] + arr[i][1] + arr[i][2] + arr[i][4] + arr[i][5] > 1 :
            res[i] = 'Two Or More'
        elif arr[i][0]  == 1:
            res[i] = 'Black/AA'
        elif arr[i][1] == 1:
            res[i] = 'Asian'
        elif arr[i][2] == 1:
            res[i] = 'Haw/Pac Isl.'
        elif arr[i][4] == 1 :
            res[i] = 'A/I AK Native'
        elif arr[i][5] == 1:
            res[i] = 'White'
        else:
            res[i] = 'Other'
            
    return res

def nb_loop(df):
    cols = [c for c in df.columns if c.startswith('eri_')]
    res = np.empty(len(df), dtype=f"<U{len('A/I AK Native')}")
    df['rno_defined'] = conditional_assignment(df[cols].values, res)
    return df

# df with 24mil rows
n = 4_000_000
df = pd.DataFrame({
    'eri_afr_amer': [0, 0, 0, 0, 0, 0]*n, 
    'eri_asian': [1, 0, 0, 0, 0, 0]*n, 
    'eri_hawaiian': [0, 0, 0, 1, 0, 0]*n, 
    'eri_hispanic': [0, 1, 0, 0, 1, 0]*n, 
    'eri_nat_amer': [0, 0, 0, 0, 1, 0]*n, 
    'eri_white': [0, 0, 1, 1, 0, 0]*n
}, dtype='int8')
df.insert(0, 'name', ['MOST', 'CRUISE', 'DEPP', 'DICAP', 'BRANDO', 'HANKS']*n)

%timeit nb_loop(df)
# 5.23 s ± 45.2 ms per loop (mean ± std. dev. of 10 runs, 10 loops each)

%timeit pd_loc(df)
# 7.97 s ± 28.8 ms per loop (mean ± std. dev. of 10 runs, 10 loops each)

%timeit np_select(df)
# 8.5 s ± 39.6 ms per loop (mean ± std. dev. of 10 runs, 10 loops each)

2:在下面的结果中，我展示了两种方法的性能，分别使用一个具有20k行和1 mil行的数据框架。对于更小的帧，间隔更小，因为当apply()是一个循环时，优化的方法有一个开销。随着帧大小的增加，向量化开销w.r.t.减少到代码的总体运行时，而apply()仍然是帧上的循环。

n = 20_000 # 1_000_000
df = pd.DataFrame(np.random.rand(n,3)-0.5, columns=['colA','colB','colC'])

%timeit df[['colA','colB']].prod(1).where(df['colC']>0, df['colA'] / df['colB'])
# n = 20000: 2.69 ms ± 23.5 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
# n = 1000000: 86.2 ms ± 441 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

%timeit df.apply(lambda row: row.colA * row.colB if row.colC > 0 else row.colA / row.colB, axis=1)
# n = 20000: 679 ms ± 33.8 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
# n = 1000000: 31.5 s ± 587 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

因为这是'pandas new column from others'的第一个谷歌结果，这里有一个简单的例子:

import pandas as pd

# make a simple dataframe
df = pd.DataFrame({'a':[1,2], 'b':[3,4]})
df
#    a  b
# 0  1  3
# 1  2  4

# create an unattached column with an index
df.apply(lambda row: row.a + row.b, axis=1)
# 0    4
# 1    6

# do same but attach it to the dataframe
df['c'] = df.apply(lambda row: row.a + row.b, axis=1)
df
#    a  b  c
# 0  1  3  4
# 1  2  4  6

---

如果你得到SettingWithCopyWarning，你也可以这样做:

fn = lambda row: row.a + row.b # define a function for the new column
col = df.apply(fn, axis=1) # get column data with an index
df = df.assign(c=col.values) # assign values to column 'c'

来源:https://stackoverflow.com/a/12555510/243392

如果你的列名包含空格，你可以使用这样的语法:

df = df.assign(**{'some column name': col.values})

---

这是apply和assign的文档。

还有另一种(易于推广的)方法，其基础是pandas.DataFrame.idxmax。首先，易于概括的序言。

# Indeed, all your conditions boils down to the following
_gt_1_key = 'two_or_more'
_lt_1_key = 'other'

# The "dictionary-based" if-else statements
labels = {
    _gt_1_key     : 'Two Or More',
    'eri_hispanic': 'Hispanic',
    'eri_nat_amer': 'A/I AK Native',
    'eri_asian'   : 'Asian',
    'eri_afr_amer': 'Black/AA',
    'eri_hawaiian': 'Haw/Pac Isl.',
    'eri_white'   : 'White',  
    _lt_1_key     : 'Other',
}

# The output-driving 1-0 matrix
mat = df.filter(regex='^eri_').copy()  # `~.copy` to avoid `SettingWithCopyWarning`

．.． 最后，以向量化的方式:

mat[_gt_1_key] = gt1 = mat.sum(axis=1)
mat[_lt_1_key] = gt1.eq(0).astype(int)
race_label     = mat.idxmax(axis=1).map(labels)

在哪里

>>> race_label
0           White
1        Hispanic
2           White
3           White
4           Other
5           White
6     Two Or More
7           White
8    Haw/Pac Isl.
9           White
dtype: object

那是一只熊猫。您可以轻松地在df中托管系列实例，即df['race_label'] = race_label。

试试这个,

df.loc[df['eri_white']==1,'race_label'] = 'White'
df.loc[df['eri_hawaiian']==1,'race_label'] = 'Haw/Pac Isl.'
df.loc[df['eri_afr_amer']==1,'race_label'] = 'Black/AA'
df.loc[df['eri_asian']==1,'race_label'] = 'Asian'
df.loc[df['eri_nat_amer']==1,'race_label'] = 'A/I AK Native'
df.loc[(df['eri_afr_amer'] + df['eri_asian'] + df['eri_hawaiian'] + df['eri_nat_amer'] + df['eri_white']) > 1,'race_label'] = 'Two Or More'
df.loc[df['eri_hispanic']==1,'race_label'] = 'Hispanic'
df['race_label'].fillna('Other', inplace=True)

O/P:

     lname   fname rno_cd  eri_afr_amer  eri_asian  eri_hawaiian  \
0      MOST    JEFF      E             0          0             0   
1    CRUISE     TOM      E             0          0             0   
2      DEPP  JOHNNY    NaN             0          0             0   
3     DICAP     LEO    NaN             0          0             0   
4    BRANDO  MARLON      E             0          0             0   
5     HANKS     TOM    NaN             0          0             0   
6    DENIRO  ROBERT      E             0          1             0   
7    PACINO      AL      E             0          0             0   
8  WILLIAMS   ROBIN      E             0          0             1   
9  EASTWOOD   CLINT      E             0          0             0   

   eri_hispanic  eri_nat_amer  eri_white rno_defined    race_label  
0             0             0          1       White         White  
1             1             0          0       White      Hispanic  
2             0             0          1     Unknown         White  
3             0             0          1     Unknown         White  
4             0             0          0       White         Other  
5             0             0          1     Unknown         White  
6             0             0          1       White   Two Or More  
7             0             0          1       White         White  
8             0             0          0       White  Haw/Pac Isl.  
9             0             0          1       White         White 

使用.loc代替apply。

它改进了向量化。

.loc的工作方式很简单，根据条件屏蔽行，对冻结行应用值。

欲了解更多细节，请访问。loc文档

性能指标:

答:接受

def label_race (row):
   if row['eri_hispanic'] == 1 :
      return 'Hispanic'
   if row['eri_afr_amer'] + row['eri_asian'] + row['eri_hawaiian'] + row['eri_nat_amer'] + row['eri_white'] > 1 :
      return 'Two Or More'
   if row['eri_nat_amer'] == 1 :
      return 'A/I AK Native'
   if row['eri_asian'] == 1:
      return 'Asian'
   if row['eri_afr_amer']  == 1:
      return 'Black/AA'
   if row['eri_hawaiian'] == 1:
      return 'Haw/Pac Isl.'
   if row['eri_white'] == 1:
      return 'White'
   return 'Other'

df=pd.read_csv('dataser.csv')
df = pd.concat([df]*1000)

%timeit df.apply(lambda row: label_race(row), axis=1)

每循环1.15 s±46.5 ms(平均±标准值7次运行，每循环1次)

我建议的答案:

def label_race(df):
    df.loc[df['eri_white']==1,'race_label'] = 'White'
    df.loc[df['eri_hawaiian']==1,'race_label'] = 'Haw/Pac Isl.'
    df.loc[df['eri_afr_amer']==1,'race_label'] = 'Black/AA'
    df.loc[df['eri_asian']==1,'race_label'] = 'Asian'
    df.loc[df['eri_nat_amer']==1,'race_label'] = 'A/I AK Native'
    df.loc[(df['eri_afr_amer'] + df['eri_asian'] + df['eri_hawaiian'] + df['eri_nat_amer'] + df['eri_white']) > 1,'race_label'] = 'Two Or More'
    df.loc[df['eri_hispanic']==1,'race_label'] = 'Hispanic'
    df['race_label'].fillna('Other', inplace=True)
df=pd.read_csv('s22.csv')
df = pd.concat([df]*1000)

%timeit label_race(df)

每循环24.7 ms±1.7 ms(平均±标准值7次运行，每循环10次)