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))
