在 Pandas DataFrame 中，我想根据另一列的值有条件地创建一个新列。在我的应用程序中，DataFrame 通常有几百万行，并且唯一条件值的数量很少，按统一顺序排列。性能极其重要:生成新列的最快方法是什么？

我在下面创建了一个示例案例，并且已经尝试并比较了不同的方法。
在示例中，条件填充表示为
基于列值的字典查找label (此处: 1, 2, 3 之一)。

lookup_dict = {
    1: 100,   # arbitrary
    2: 200,   # arbitrary
    3: 300,   # arbitrary
    }

       label  output
0      3     300
1      2     200
2      3     300
3      3     300
4      2     200
5      2     200
6      1     100
7      1     100

def fill_output(r):
    ''' called by groupby().apply(): all r.label values are the same '''
    r.loc[:, 'output'] = lookup_dict[r.iloc[0]['label']]
    return r

df = df.groupby('label').apply(fill_output)

>>> method_1_groupby ran in 2.29s (average over 3 iterations)

dgb = df.groupby('label')
for label, idx in dgb.indices.items():
    df.loc[idx, 'output'] = lookup_dict[label]

method_2_indices ran in 1.21s (average over 3 iterations)

df['output'] = df.label.map(lookup_dict.get)

for label, value in lookup_dict.items():
    df.loc[df.label == label, 'output'] = value

method_4_forloop ran in 0.54s (average over 3 iterations)

conditions = [df.label == k for k in lookup_dict.keys()]
choices = list(lookup_dict.values())

df['output'] = np.select(conditions, choices)

method_5_select ran in 0.29s (average over 3 iterations)

@jit(int64[:](int64[:]), nopython=True)
def hardcoded_conditional_filling(column):
    output = np.zeros_like(column)
    i = 0
    for c in column:
        if c == 1:
            output[i] = 100
        elif c == 2:
            output[i] = 200
        elif c == 3:
            output[i] = 300
        i += 1
    return output

df['output'] = hardcoded_conditional_filling(df.label.values)

method_6_numba ran in 0.19s (average over 3 iterations)

import pandas as pd
import numpy as np
from timeit import timeit
from numba import jit, int64

lookup_dict = {
        1: 100,   # arbitrary
        2: 200,   # arbitrary
        3: 300,   # arbitrary
        }

Nlines = int(1e7)

# Generate 
label = np.round(np.random.rand(Nlines)*2+1).astype(np.int64)
df0 = pd.DataFrame(label, columns=['label'])

# Now the goal is to assign the look_up_dict values to a new column 'output' 
# based on the value of label

# Method 1
# using groupby().apply()

def method_1_groupby(df):

    def fill_output(r):
        ''' called by groupby().apply(): all r.label values are the same '''
        #print(r.iloc[0]['label'])   # activate to reveal the #2936 issue in Pandas
        r.loc[:, 'output'] = lookup_dict[r.iloc[0]['label']]
        return r

    df = df.groupby('label').apply(fill_output)
    return df 

def method_2_indices(df):

    dgb = df.groupby('label')
    for label, idx in dgb.indices.items():
        df.loc[idx, 'output'] = lookup_dict[label]

    return df

def method_3_map(df):

    df['output'] = df.label.map(lookup_dict.get)

    return df

def method_4_forloop(df):
    ''' naive '''

    for label, value in lookup_dict.items():
        df.loc[df.label == label, 'output'] = value

    return df

def method_5_select(df):
    ''' Based on answer from 
    https://stackoverflow.com/a/19913845/5622825
    '''

    conditions = [df.label == k for k in lookup_dict.keys()]
    choices = list(lookup_dict.values())

    df['output'] = np.select(conditions, choices)

    return df

def method_6_numba(df):
    ''' This works, but it is hardcoded and i don't really know how
    to make it compile with list as runtime constants'''


    @jit(int64[:](int64[:]), nopython=True)
    def hardcoded_conditional_filling(column):
        output = np.zeros_like(column)
        i = 0
        for c in column:
            if c == 1:
                output[i] = 100
            elif c == 2:
                output[i] = 200
            elif c == 3:
                output[i] = 300
            i += 1
        return output

    df['output'] = hardcoded_conditional_filling(df.label.values)

    return df

df1 = method_1_groupby(df0)
df2 = method_2_indices(df0.copy())
df3 = method_3_map(df0.copy())
df4 = method_4_forloop(df0.copy())
df5 = method_5_select(df0.copy())
df6 = method_6_numba(df0.copy())

# make sure we havent modified the input (would bias the results)
assert 'output' not in df0.columns 

# Test validity
assert (df1 == df2).all().all()
assert (df1 == df3).all().all()
assert (df1 == df4).all().all()
assert (df1 == df5).all().all()
assert (df1 == df6).all().all()

# Compare performances
Nites = 3
print('Compare performances for {0:.1g} lines'.format(Nlines))
print('-'*30)
for method in [
               'method_1_groupby', 'method_2_indices', 
               'method_3_map', 'method_4_forloop', 
               'method_5_select', 'method_6_numba']:
    print('{0} ran in {1:.2f}s (average over {2} iterations)'.format(
            method, 
            timeit("{0}(df)".format(method), setup="from __main__ import df0, {0}; df=df0.copy()".format(method), number=Nites)/Nites,
            Nites))

Compare performances for 1e+07 lines
------------------------------
method_1_groupby ran in 2.29s (average over 3 iterations)
method_2_indices ran in 1.21s (average over 3 iterations)
method_3_map ran in 3.07s (average over 3 iterations)
method_4_forloop ran in 0.54s (average over 3 iterations)
method_5_select ran in 0.29s (average over 3 iterations)
method_6_numba ran in 0.19s (average over 3 iterations)

def method_3b_mapdirect(df):
    ''' Suggested by https://stackoverflow.com/a/51388828/5622825'''

    df['output'] = df.label.map(lookup_dict)

    return df

def method_7_take(df):
    ''' Based on answer from 
    https://stackoverflow.com/a/19913845/5622825

    Exploiting that labels are continuous integers
    '''

    lookup_arr = np.array(list(lookup_dict.values()))
    df['output'] = lookup_arr.take(df['label'] - 1)

    return df

method_3_mapdirect ran in 0.23s (average over 3 iterations)
method_7_take ran in 0.11s (average over 3 iterations)

最佳答案

我会考虑 .map (#3) 这样做的惯用方法 - 但不要通过 .get - 单独使用字典，应该会看到非常显着的改进。

df = pd.DataFrame({'label': np.random.randint(, 4, size=1000000, dtype='i8')})

%timeit df['output'] = df.label.map(lookup_dict.get)
261 ms ± 12.7 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

%timeit df['output'] = df.label.map(lookup_dict)
69.6 ms ± 3.08 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

%%timeit
lookup_arr = np.array(list(lookup_dict.values()))
df['output'] = lookup_arr.take(df['label'] - 1)
8.68 ms ± 332 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

关于python - 有条件地创建 Pandas 列的最快方法，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/51388201/