好的，有两个步骤--第一步是编写一个函数来完成你想要的转换--我已经根据你的伪代码整理了一个例子：

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'

您可能想过一遍，但它似乎起到了作用-注意，进入函数的参数被认为是一个标记为“行”的Series对象。
接下来，在Pandas中使用apply函数来应用函数-例如

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

注意axis=1说明符，这意味着应用程序是在行级别而不是列级别执行的。结果如下所示：

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

如果您对这些结果感到满意，那么再次运行它，将结果保存到原始 Dataframe 的新列中。

df['race_label'] = df.apply (lambda row: label_race(row), axis=1)

生成的 Dataframe 如下所示（向右滚动以查看新列）：

lname   fname rno_cd  eri_afr_amer  eri_asian  eri_hawaiian   eri_hispanic  eri_nat_amer  eri_white rno_defined    race_label
0      MOST    JEFF      E             0          0             0              0             0          1       White         White
1    CRUISE     TOM      E             0          0             0              1             0          0       White      Hispanic
2      DEPP  JOHNNY    NaN             0          0             0              0             0          1     Unknown         White
3     DICAP     LEO    NaN             0          0             0              0             0          1     Unknown         White
4    BRANDO  MARLON      E             0          0             0              0             0          0       White         Other
5     HANKS     TOM    NaN             0          0             0              0             0          1     Unknown         White
6    DENIRO  ROBERT      E             0          1             0              0             0          1       White   Two Or More
7    PACINO      AL      E             0          0             0              0             0          1       White         White
8  WILLIAMS   ROBIN      E             0          0             1              0             0          0       White  Haw/Pac Isl.
9  EASTWOOD   CLINT      E             0          0             0              0             0          1       White         White

由于这是第一个谷歌搜索结果'Pandas新专栏从其他'，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

如果你得到了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
如果列名中包含空格，则可以使用如下语法：

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

这是apply和assign的文档。

上面的答案是完全正确的，但是存在一个矢量化的解决方案，其形式为numpy.select。这允许您定义条件，然后定义这些条件的输出，这比使用apply要高效得多：
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

为什么要使用numpy.select而不是apply？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给我们带来了“巨大”的性能改进，而且这种差异只会随着数据的增长而增加。

.apply()取函数作为第一个参数;传入label_race函数，如下所示：

df['race_label'] = df.apply(label_race, axis=1)

你不需要创建一个lambda函数来传入一个函数。

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。
它改进了矢量化。
.loc的工作方式很简单，根据条件屏蔽行，将值应用于冻结行。
有关详细信息，请访问.loc docs
性能指标：
接受的答案：

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)

1.15 s ± 46.5 ms/循环（7次运行的平均值±标准差，每次运行1个循环）
我的建议答案：

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)

24.7 ms ± 1.7 ms/循环（7次运行的平均值±标准差，每次运行10次循环）

如果我们检查它的source code，apply()是Python for循环的语法糖（通过FrameApply类的apply_series_generator()方法），因为它有panda开销，它通常比Python循环"慢"。
尽可能使用优化的（矢量化的）方法。如果你必须使用循环，使用@numba.jit装饰器。

1.不要将apply()用于if-else阶梯

df.apply()几乎是Pandas中最慢的方法。如user3483203和Mohamed Thasin ah的答案所示，根据 Dataframe 大小，np.select()和df.loc可能比df.apply()快50 - 300倍才能产生相同的输出。
实际上，使用numba模块中的@jit装饰器的循环实现（与apply()类似）比df.loc和np.select快（大约50 - 60%）。1
Numba可以处理numpy数组，所以在使用jit装饰器之前，需要将 Dataframe 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，你必须通过循环中的索引来访问这些列。if- jitted函数中的else ladders在apply()中的else ladders上是通过索引访问列的，除此之外几乎是相同的实现。

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()进行数值运算

如果需要通过添加两列来添加新行，您的第一React可能是编写

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

但是，使用sum(axis=1)方法（如果只有几列，则使用+运算符）执行按行加法：

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

根据 Dataframe 大小，sum(1)可能比apply()快100倍。
事实上，在panda Dataframe 上，几乎不需要使用apply()进行数值运算，因为它对大多数运算都进行了优化：加法（sum(1)），减法（sub()或diff()），乘法（prod(1)），部门（1个一米26英寸1 x或1个一米27英寸1 x），电源（pow()）、>、>=、==、%、//、&、|等都可以在没有apply()的整个 Dataframe 上执行。
例如，假设您要使用以下规则创建一个新列：

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

使用优化的panda方法，这可以写成

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

等效的apply()溶液为：

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

对于20k行的 Dataframe ，使用优化方法的方法比等效的apply()方法快250倍。这个差距只会随着数据大小的增加而增加（对于1mil行的 Dataframe ，它快365倍），时间差异将变得越来越明显。2
1：在下面的结果中，我使用24mil行的 Dataframe （这是我可以在我的机器上构建的最大帧）展示了三种方法的性能。对于较小的帧，numba-jitted函数始终比其他两种方法至少快50%（您可以自己检查）。

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)

二：在下面的结果中，我使用20k行和1mil行的 Dataframe 显示了两种方法的性能。对于较小的帧，差距较小，因为优化的方法有开销，而apply()是一个循环。随着帧大小的增加，矢量化开销成本减少了对代码的整个运行时间的影响，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)

As @user3483203 pointed out, numpy.select is the best approach
将条件语句和相应的操作存储在两个列表中

conds = [(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,])

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

You can now use np.select using these lists as its arguments

df['label_race'] = np.select(conds,actions,default='Other')

参考：https://numpy.org/doc/stable/reference/generated/numpy.select.html

还有另一种（易于推广的）方法，其基石是pandas.DataFrame.idxmax。

# 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，在 vectorized fashion 中：

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)

其中

>>> 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

这是一个pandas.Series示例，您可以轻松地将其托管在df中，即执行df['race_label'] = race_label。