You can go 40+ times faster than the top answers here if you do your math in numpy instead. Adapting @Rocky K's top two answers. The main difference is running on an actual df of 120k rows. Numpy is way faster at math when you apply your functions array-wise (instead of applying a function value-wise). The best answer is by far the third one because it uses numpy for the math. Also notice that it only calculates 1024**2 and 1024**3 once each instead of once for each row, saving 240k calculations. Here are the timings on my machine:

Tuples (pass value, return tuple then zip, new columns dont exist):
Runtime: 10.935037851333618 

Tuples (pass value, return tuple then zip, new columns exist):
Runtime: 11.120025157928467 

Use numpy for math portions:
Runtime: 0.24799370765686035

以下是我用来计算这些时间的脚本(改编自Rocky K):

import numpy as np
import pandas as pd
import locale
import time

size = np.random.random(120000) * 1000000000
data = pd.DataFrame({'Size': size})

def sizes_pass_value_return_tuple(value):
    a = locale.format_string("%.1f", value / 1024.0, grouping=True) + ' KB'
    b = locale.format_string("%.1f", value / 1024.0 ** 2, grouping=True) + ' MB'
    c = locale.format_string("%.1f", value / 1024.0 ** 3, grouping=True) + ' GB'
    return a, b, c

print('\nTuples (pass value, return tuple then zip, new columns dont exist):')
df1 = data.copy()
start = time.time()
df1['size_kb'],  df1['size_mb'], df1['size_gb'] = zip(*df1['Size'].apply(sizes_pass_value_return_tuple))
end = time.time()
print('Runtime:', end - start, '\n')

print('Tuples (pass value, return tuple then zip, new columns exist):')
df2 = data.copy()
start = time.time()
df2 = pd.concat([df2, pd.DataFrame(columns=['size_kb', 'size_mb', 'size_gb'])])
df2['size_kb'],  df2['size_mb'], df2['size_gb'] = zip(*df2['Size'].apply(sizes_pass_value_return_tuple))
end = time.time()
print('Runtime:', end - start, '\n')

print('Use numpy for math portions:')
df3 = data.copy()
start = time.time()
df3['size_kb'] = (df3.Size.values / 1024).round(1)
df3['size_kb'] = df3.size_kb.astype(str) + ' KB'
df3['size_mb'] = (df3.Size.values / 1024 ** 2).round(1)
df3['size_mb'] = df3.size_mb.astype(str) + ' MB'
df3['size_gb'] = (df3.Size.values / 1024 ** 3).round(1)
df3['size_gb'] = df3.size_gb.astype(str) + ' GB'
end = time.time()
print('Runtime:', end - start, '\n')

顶部答案之间的性能差异很大，Jesse和famaral42已经讨论过这个问题，但是值得分享顶部答案之间的公平比较，并详细说明Jesse答案中一个微妙但重要的细节:传递给函数的参数也会影响性能。

(Python 3.7.4, Pandas 1.0.3)

import pandas as pd
import locale
import timeit


def create_new_df_test():
    df_test = pd.DataFrame([
      {'dir': '/Users/uname1', 'size': 994933},
      {'dir': '/Users/uname2', 'size': 109338711},
    ])
    return df_test


def sizes_pass_series_return_series(series):
    series['size_kb'] = locale.format_string("%.1f", series['size'] / 1024.0, grouping=True) + ' KB'
    series['size_mb'] = locale.format_string("%.1f", series['size'] / 1024.0 ** 2, grouping=True) + ' MB'
    series['size_gb'] = locale.format_string("%.1f", series['size'] / 1024.0 ** 3, grouping=True) + ' GB'
    return series


def sizes_pass_series_return_tuple(series):
    a = locale.format_string("%.1f", series['size'] / 1024.0, grouping=True) + ' KB'
    b = locale.format_string("%.1f", series['size'] / 1024.0 ** 2, grouping=True) + ' MB'
    c = locale.format_string("%.1f", series['size'] / 1024.0 ** 3, grouping=True) + ' GB'
    return a, b, c


def sizes_pass_value_return_tuple(value):
    a = locale.format_string("%.1f", value / 1024.0, grouping=True) + ' KB'
    b = locale.format_string("%.1f", value / 1024.0 ** 2, grouping=True) + ' MB'
    c = locale.format_string("%.1f", value / 1024.0 ** 3, grouping=True) + ' GB'
    return a, b, c

以下是调查结果:

# 1 - Accepted (Nels11 Answer) - (pass series, return series):
9.82 ms ± 377 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

# 2 - Pandafied (jaumebonet Answer) - (pass series, return tuple):
2.34 ms ± 48.6 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

# 3 - Tuples (pass series, return tuple then zip):
1.36 ms ± 62.8 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

# 4 - Tuples (Jesse Answer) - (pass value, return tuple then zip):
752 µs ± 18.5 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

请注意，返回元组是最快的方法，但作为参数传入的内容也会影响性能。代码中的差异很小，但性能的提高是显著的。

测试#4(传入单个值)的速度是测试#3(传入一系列值)的两倍，尽管执行的操作表面上是相同的。

但还有更多……

# 1a - Accepted (Nels11 Answer) - (pass series, return series, new columns exist):
3.23 ms ± 141 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

# 2a - Pandafied (jaumebonet Answer) - (pass series, return tuple, new columns exist):
2.31 ms ± 39.3 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

# 3a - Tuples (pass series, return tuple then zip, new columns exist):
1.36 ms ± 58.4 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

# 4a - Tuples (Jesse Answer) - (pass value, return tuple then zip, new columns exist):
694 µs ± 3.9 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

在某些情况下(#1a和#4a)，将函数应用到已经存在输出列的DataFrame比从函数中创建输出列更快。

下面是运行测试的代码:

# Paste and run the following in ipython console. It will not work if you run it from a .py file.
print('\nAccepted Answer (pass series, return series, new columns dont exist):')
df_test = create_new_df_test()
%timeit result = df_test.apply(sizes_pass_series_return_series, axis=1)
print('Accepted Answer (pass series, return series, new columns exist):')
df_test = create_new_df_test()
df_test = pd.concat([df_test, pd.DataFrame(columns=['size_kb', 'size_mb', 'size_gb'])])
%timeit result = df_test.apply(sizes_pass_series_return_series, axis=1)

print('\nPandafied (pass series, return tuple, new columns dont exist):')
df_test = create_new_df_test()
%timeit df_test[['size_kb', 'size_mb', 'size_gb']] = df_test.apply(sizes_pass_series_return_tuple, axis=1, result_type="expand")
print('Pandafied (pass series, return tuple, new columns exist):')
df_test = create_new_df_test()
df_test = pd.concat([df_test, pd.DataFrame(columns=['size_kb', 'size_mb', 'size_gb'])])
%timeit df_test[['size_kb', 'size_mb', 'size_gb']] = df_test.apply(sizes_pass_series_return_tuple, axis=1, result_type="expand")

print('\nTuples (pass series, return tuple then zip, new columns dont exist):')
df_test = create_new_df_test()
%timeit df_test['size_kb'],  df_test['size_mb'], df_test['size_gb'] = zip(*df_test.apply(sizes_pass_series_return_tuple, axis=1))
print('Tuples (pass series, return tuple then zip, new columns exist):')
df_test = create_new_df_test()
df_test = pd.concat([df_test, pd.DataFrame(columns=['size_kb', 'size_mb', 'size_gb'])])
%timeit df_test['size_kb'],  df_test['size_mb'], df_test['size_gb'] = zip(*df_test.apply(sizes_pass_series_return_tuple, axis=1))

print('\nTuples (pass value, return tuple then zip, new columns dont exist):')
df_test = create_new_df_test()
%timeit df_test['size_kb'],  df_test['size_mb'], df_test['size_gb'] = zip(*df_test['size'].apply(sizes_pass_value_return_tuple))
print('Tuples (pass value, return tuple then zip, new columns exist):')
df_test = create_new_df_test()
df_test = pd.concat([df_test, pd.DataFrame(columns=['size_kb', 'size_mb', 'size_gb'])])
%timeit df_test['size_kb'],  df_test['size_mb'], df_test['size_gb'] = zip(*df_test['size'].apply(sizes_pass_value_return_tuple))

我相信1.1版本打破了上面答案中建议的行为。

import pandas as pd
def test_func(row):
    row['c'] = str(row['a']) + str(row['b'])
    row['d'] = row['a'] + 1
    return row

df = pd.DataFrame({'a': [1, 2, 3], 'b': ['i', 'j', 'k']})
df.apply(test_func, axis=1)

上面的代码在pandas 1.1.0上运行返回:

   a  b   c  d
0  1  i  1i  2
1  1  i  1i  2
2  1  i  1i  2

而在熊猫1.0.5中，它返回:

   a   b    c  d
0  1   i   1i  2
1  2   j   2j  3
2  3   k   3k  4

我想这是你所期望的。

不确定发布说明如何解释这种行为，但是正如这里所解释的那样，通过复制原始行来避免突变，从而恢复旧的行为。例如:

def test_func(row):
    row = row.copy()   #  <---- Avoid mutating the original reference
    row['c'] = str(row['a']) + str(row['b'])
    row['d'] = row['a'] + 1
    return row

这是一种非常快速的方法，用apply和。只需将多个值作为列表返回，然后使用to_list()

import pandas as pd

dat = [ [i, 10*i] for i in range(100000)]

df = pd.DataFrame(dat, columns = ["a","b"])

def add_and_div(x):
    add = x + 3
    div = x / 3
    return [add, div]

start = time.time()
df[['c','d']] = df['a'].apply(lambda x: add_and_div(x)).to_list()
end = time.time()

print(end-start) # output: 0.27606

它提供了一个新的数据框架，其中包含原始数据框架的两列。

import pandas as pd
df = ...
df_with_two_columns = df.apply(lambda row:pd.Series([row['column_1'], row['column_2']], index=['column_1', 'column_2']),axis = 1)

非常酷的答案!谢谢Jesse和jaumebonet!以下是我对以下方面的一些观察:

邮政编码(*…… ．.． result_type = "扩大")

虽然expand更优雅(pandifyed)，但**zip至少快2倍。在下面这个简单的例子中，我的速度快了4倍。

import pandas as pd

dat = [ [i, 10*i] for i in range(1000)]

df = pd.DataFrame(dat, columns = ["a","b"])

def add_and_sub(row):
    add = row["a"] + row["b"]
    sub = row["a"] - row["b"]
    return add, sub

df[["add", "sub"]] = df.apply(add_and_sub, axis=1, result_type="expand")
# versus
df["add"], df["sub"] = zip(*df.apply(add_and_sub, axis=1))