最快的选项不仅取决于 DataFrame 的长度(在本例中约为 13M 行),还取决于组的数量。下面的性能图比较了寻找每组最大值的多种方法:
If there an only a few (large) groups, using_idxmax
may be the fastest option:
If there are many (small) groups and the DataFrame is not too large, using_sort_drop
may be the fastest option:
但请记住,虽然using_sort_drop
, using_sort
and using_rank
开始看起来很快,因为N = len(df)
增加,它们相对于其他选项的速度很快消失。对于足够大的N
, using_idxmax
成为最快的选择,即使有很多组。
using_sort_drop
, using_sort
and using_rank
对 DataFrame(或 DataFrame 中的组)进行排序。排序是O(N * log(N))
平均而言,而其他方法使用O(N)
运营。这就是为什么像这样的方法using_idxmax
beats using_sort_drop
对于非常大的数据框。
请注意,基准测试结果可能会因多种原因而有所不同,包括机器规格、操作系统和软件版本。因此,在您自己的机器上运行基准测试并使用适合您情况的测试数据非常重要。
基于上面的性能图,using_sort_drop
may be对于 13M 行的 DataFrame,这是一个值得考虑的选项,特别是如果它有许多(小)组。不然我会怀疑using_idxmax
成为最快的选择——但同样,检查机器上的基准测试也很重要。
这是我用来制作的设置性能图 https://github.com/nschloe/perfplot:
import numpy as np
import pandas as pd
import perfplot
def make_df(N):
# lots of small groups
df = pd.DataFrame(np.random.randint(N//10+1, size=(N, 2)), columns=['Id','delta'])
# few large groups
# df = pd.DataFrame(np.random.randint(10, size=(N, 2)), columns=['Id','delta'])
return df
def using_idxmax(df):
return df.loc[df.groupby("Id")['delta'].idxmax()]
def max_mask(s):
i = np.asarray(s).argmax()
result = [False]*len(s)
result[i] = True
return result
def using_custom_mask(df):
mask = df.groupby("Id")['delta'].transform(max_mask)
return df.loc[mask]
def using_isin(df):
idx = df.groupby("Id")['delta'].idxmax()
mask = df.index.isin(idx)
return df.loc[mask]
def using_sort(df):
df = df.sort_values(by=['delta'], ascending=False, kind='mergesort')
return df.groupby('Id', as_index=False).first()
def using_rank(df):
mask = (df.groupby('Id')['delta'].rank(method='first', ascending=False) == 1)
return df.loc[mask]
def using_sort_drop(df):
# Thanks to jezrael
# https://stackoverflow.com/questions/50381064/select-the-max-row-per-group-pandas-performance-issue/50389889?noredirect=1#comment87795818_50389889
return df.sort_values(by=['delta'], ascending=False, kind='mergesort').drop_duplicates('Id')
def using_apply(df):
selected_idx = df.groupby("Id").apply(lambda df: df.delta.argmax())
return df.loc[selected_idx]
def check(df1, df2):
df1 = df1.sort_values(by=['Id','delta'], kind='mergesort').reset_index(drop=True)
df2 = df2.sort_values(by=['Id','delta'], kind='mergesort').reset_index(drop=True)
return df1.equals(df2)
perfplot.show(
setup=make_df,
kernels=[using_idxmax, using_custom_mask, using_isin, using_sort,
using_rank, using_apply, using_sort_drop],
n_range=[2**k for k in range(2, 20)],
logx=True,
logy=True,
xlabel='len(df)',
repeat=75,
equality_check=check)
另一种基准测试方法是使用IPython %timeit https://stackoverflow.com/a/29280612/190597:
In [55]: df = make_df(2**20)
In [56]: %timeit using_sort_drop(df)
1 loop, best of 3: 403 ms per loop
In [57]: %timeit using_rank(df)
1 loop, best of 3: 1.04 s per loop
In [58]: %timeit using_idxmax(df)
1 loop, best of 3: 15.8 s per loop