pandas.core.groupby.GroupBy.agg 是按组计算多列的有效解决方案吗?

Question

假设你有一个这样的pandas.DataFrame：

import pandas as pd
import numpy as np
from math import exp, log10
df = pd.DataFrame({'S' : ['C', 'A', 'A', 'B', 'B', 'A'],
'ID': ['ID3', 'ID1', 'ID2', 'ID4', 'ID4', 'ID1'],
'M' : ['Y', 'X', 'X', 'Y', 'Z', 'X'],                   
'V' : ['<1', '<0.5', 3, '>10', 7, 6]
})
df
#   S   ID  M     V
#0  C  ID3  Y    <1
#1  A  ID1  X  <0.5
#2  A  ID2  X     3
#3  B  ID4  Y   >10
#4  B  ID4  Z     7
#5  A  ID1  X     6

以及将每个M映射到 x 函数的字典，例如：

M_to_transf_dict = {'X' : 'x', 'Y' : 'exp(x)', 'Z' : 'log10(x)'}

这是您需要做的：

• 将V拆分为数字部分(V_u)和限定符(V_c："<"，">"，如果尚未成为数字或可解释)
• 按S, M分组，对于每个组：
  • 计算V_u的平均V_mean、V_N计数和样本标准差V_sample_sd
  • 制作唯一ID的逗号分隔列表，并将结果存储到ID
  • 查找V_c中最常用的"<"或">"限定符，并将结果存储到V_mean_c
  • 应用每个M对应的x函数进行V_mean，并将结果存储到TV_mean
  • 在"<"或">"时V_mean和TV_mean上应用V_mean_c

经过多次尝试，我整理了一些似乎有效的东西，但我对性能有疑问.
我看到一些帖子(例如这个)质疑.apply的使用，事实上，从一些测试来看，将V拆分为外部映射后进行列分配比.apply快得多：

def unc(v):
try:
u = float(v)
c = "="
except ValueError:
v = str(v)
u = float(v[1:])
c = v[0]
except:
u = pd.NA
c = "="
return [u, c]
%timeit df[['V_u', 'V_c']] = df.apply(lambda row : unc(row['V']), axis = 1, result_type = 'expand')
# 698 µs ± 5.22 µs per loop (mean ± std. dev. of 7 runs, 1,000 loops each)
%%timeit
uc = df['V'].map(unc)
u = [uci[0] for uci in uc.values]
c = [uci[1] for uci in uc.values]
df['V_u'] = u
df['V_c'] = c
# 129 µs ± 3.59 µs per loop (mean ± std. dev. of 7 runs, 10,000 loops each)
%%timeit
uc = df['V'].map(unc)
u, c = [], []
for uci in uc:
u.append(uci[0])
c.append(uci[1])
df['V_u'] = u
df['V_c'] = c
# 124 µs ± 1.11 µs per loop (mean ± std. dev. of 7 runs, 10,000 loops each)

(请注意，我在每次timeit之前都重制了df)。

对于我描述的其他操作，除了最后两个操作，我使用了带有命名聚合的.groupby.agg：

def majority(c_list):
c_list = list(c_list)
Nlt = c_list.count('<')
Ngt = c_list.count('>')
if Nlt + Ngt == 0:
c = '='
else:
c = sorted(zip([Nlt, Ngt], ['<','>']))[1][1]
return c
%%timeit
df_summary_by_S_M = df.groupby(['S','M'], as_index = False).agg(
ID = pd.NamedAgg(column = 'ID', aggfunc = lambda x : ','.join(np.unique(x))),
V_mean = pd.NamedAgg(column = 'V_u', aggfunc = 'mean'),
V_mean_c = pd.NamedAgg(column = 'V_c', aggfunc = majority),
V_N = pd.NamedAgg(column = 'V_u', aggfunc = 'count'),
V_sample_sd = pd.NamedAgg(column = 'V_u', aggfunc = 'std')
)
# 4.17 ms ± 61.1 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
df_summary_by_S_M
#   S  M       ID     V_mean V_mean_c  V_N  V_sample_sd
#0  A  X  ID1,ID2   3.166667        <    3     2.753785
#1  B  Y      ID4  10.000000        >    1          NaN
#2  B  Z      ID4   7.000000        =    1          NaN
#3  C  Y      ID3   1.000000        <    1          NaN

我不知道如何放置这个时间，特别是因为我想.agg是一种.apply形式，所以我可能没有使用最好的方法，看过之前的结果.
在我正在处理的真实数据集上，大约有 450 K 条记录，通过分组减少到 230 K， 此聚合大约需要 1 分钟.
我将不得不处理更大的数据集，这可能会成为一个问题。

有人能够建议一种更有效/性能更高的方法来进行我描述的分组计算吗？
或者这被认为是这种操作的最新性能吗？我真的没有基准来判断这一点，因此我在这里提出问题。

对于最后两个步骤，我想我将使用列表推导，例如：

%timeit df_summary_by_S_M['TV_mean'] = [eval(t) for x, t in 
zip(df_summary_by_S_M['V_mean'], df_summary_by_S_M['M'].map(M_to_transf_dict))]
# 326 µs ± 5.06 µs per loop (mean ± std. dev. of 7 runs, 1,000 loops each)
df_summary_by_S_M
#   S  M       ID     V_mean V_mean_c  V_N  V_sample_sd       TV_mean
#0  A  X  ID1,ID2   3.166667        <    3     2.753785      3.166667
#1  B  Y      ID4  10.000000        >    1          NaN  22026.465795
#2  B  Z      ID4   7.000000        =    1          NaN      0.845098
#3  C  Y      ID3   1.000000        <    1          NaN      2.718282

和：

df_summary_by_S_M['V_mean'] = [c + str(v) if c != '=' else v for c, v in 
zip(df_summary_by_S_M['V_mean_c'], df_summary_by_S_M['V_mean'])]
df_summary_by_S_M['TV_mean'] = [c + str(v) if c != '=' else v for c, v in 
zip(df_summary_by_S_M['V_mean_c'], df_summary_by_S_M['TV_mean'])]
df_summary_by_S_M
#   S  M       ID               V_mean V_mean_c  V_N  V_sample_sd  
#0  A  X  ID1,ID2  <3.1666666666666665        <    3     2.753785   
#1  B  Y      ID4                >10.0        >    1          NaN   
#2  B  Z      ID4                  7.0        =    1          NaN   
#3  C  Y      ID3                 <1.0        <    1          NaN   
#               TV_mean  
#0  <3.1666666666666665  
#1  >22026.465794806718  
#2             0.845098  
#3   <2.718281828459045  

同样，除非有人可以提出更有效的替代方案(？

>EDIT试用@mozway的代码

原始代码，所有操作一起计时。

import pandas as pd
import numpy as np
from math import exp, log10
df0 = pd.DataFrame({'S' : ['C', 'A', 'A', 'B', 'B', 'A'],
'ID': ['ID3', 'ID1', 'ID2', 'ID4', 'ID4', 'ID1'],
'M' : ['Y', 'X', 'X', 'Y', 'Z', 'X'],                   
'V' : ['<1', '<0.5', 3, '>10', 7, 6]
})
M_to_transf_dict = {'X' : 'x', 'Y' : 'exp(x)', 'Z' : 'log10(x)'}
def unc(v):
try:
u = float(v)
c = "="
except ValueError:
v = str(v)
u = float(v[1:])
c = v[0]
except:
u = pd.NA
c = "="
return [u, c]
def majority(c_list):
c_list = list(c_list)
Nlt = c_list.count('<')
Ngt = c_list.count('>')
if Nlt + Ngt == 0:
c = '='
else:
c = sorted(zip([Nlt, Ngt], ['<','>']))[1][1]
return c
%%timeit
df = df0.copy()
uc = df['V'].map(unc)
u, c = [], []
for uci in uc:
u.append(uci[0])
c.append(uci[1])
df['V_u'] = u
df['V_c'] = c
df = df.groupby(['S','M'], as_index = False).agg(
ID = pd.NamedAgg(column = 'ID', aggfunc = lambda x : ','.join(np.unique(x))),
V_mean = pd.NamedAgg(column = 'V_u', aggfunc = 'mean'),
V_mean_c = pd.NamedAgg(column = 'V_c', aggfunc = majority),
V_N = pd.NamedAgg(column = 'V_u', aggfunc = 'count'),
V_sample_sd = pd.NamedAgg(column = 'V_u', aggfunc = 'std')
)
df['TV_mean'] = [eval(t) for x, t in 
zip(df['V_mean'], df['M'].map(M_to_transf_dict))]
df['V_mean'] = [c + str(v) if c != '=' else v for c, v in 
zip(df['V_mean_c'], df['V_mean'])]
df['TV_mean'] = [c + str(v) if c != '=' else v for c, v in 
zip(df['V_mean_c'], df['TV_mean'])]

结果：

# 5.29 ms ± 100 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

@mozway的代码。

def majority(s):
mode = s.mode()
return '=' if len(mode)>1 else mode.iloc[0]
M_dic = {'Y' : np.exp, 'Z' : np.log10}
df = df0.copy()
%%timeit
(df
.join(df['V']
.str.extract('(?P<V_c>D)?(?P<V_u>d+(?:.d+)?)')
.astype({'V_u': float}).fillna({'V_c': '=', 'V_u': df['V']})
)
.assign(TV=lambda d: d.groupby('M')['V_u'].apply(lambda g: M_dic[g.name](g) if g.name in M_dic else g))
.groupby(['S','M'], as_index = False)
.agg(**{'ID': ('ID', lambda x: ','.join(x.unique())),
'V_mean': ('V_u', 'mean'),
'V_mean_c': ('V_c', majority), ## FIXME
'V_N': ('V_u', 'count'),
'V_sample_sd': ('V_u', 'std'),
'TV_mean': ('TV', 'mean'),
})
.assign(TV_mean=lambda d: d['V_mean_c'].mask(d['V_mean_c'].eq('='), '')+d['TV_mean'].astype(str))
)

结果：

# 8.05 ms ± 259 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

---

EDIT 2在更逼真的模拟数据集上重试

import pandas as pd
import numpy as np
from numpy.random import default_rng
from math import exp, log10
# Simulate dataset to process
rng = default_rng(12345)
S = rng.choice(range(250000), 450000)
ID = S.copy()
S = [f"S_{Si}" for Si in S]
ID = [f"ID_{IDi}" for IDi in ID]
pM = [np.sqrt(1+i) for i in range(100)]
pM = pM / np.sum(pM)
M = rng.choice(range(100), 450000, p = pM)
M_to_transf_dict = dict()
for i in range(0,10):
M_to_transf_dict[f"M_{i}"] = 'exp(x)'
for i in range(10,30):
M_to_transf_dict[f"M_{i}"] = 'x'
for i in range(30,100):
M_to_transf_dict[f"M_{i}"] = 'log10(x)'
M = [f"M_{Mi}" for Mi in M]
V = rng.random(450000)
Q = rng.choice(['', '<', '>'], 450000, p = [0.9, 0.05, 0.05])
V = [f"{q}{v}" for q, v in zip(Q, V)]
df0 = pd.DataFrame({'S' : S, 'ID' : ID, 'M' : M, 'V' : V})

原始代码：

def unc(v):
try:
u = float(v)
c = "="
except ValueError:
v = str(v)
u = float(v[1:])
c = v[0]
except:
u = pd.NA
c = "="
return [u, c]
def majority(c_list):
c_list = list(c_list)
Nlt = c_list.count('<')
Ngt = c_list.count('>')
if Nlt + Ngt == 0:
c = '='
else:
c = sorted(zip([Nlt, Ngt], ['<','>']))[1][1]
return c
%%timeit
df = df0.copy()
uc = df['V'].map(unc)
u, c = [], []
for uci in uc:
u.append(uci[0])
c.append(uci[1])
df['V_u'] = u
df['V_c'] = c
df = df.groupby(['S','M'], as_index = False).agg(
ID = pd.NamedAgg(column = 'ID', aggfunc = lambda x : ','.join(np.unique(x))),
V_mean = pd.NamedAgg(column = 'V_u', aggfunc = 'mean'),
V_mean_c = pd.NamedAgg(column = 'V_c', aggfunc = majority),
V_N = pd.NamedAgg(column = 'V_u', aggfunc = 'count'),
V_sample_sd = pd.NamedAgg(column = 'V_u', aggfunc = 'std')
)
df['TV_mean'] = [eval(t) for x, t in 
zip(df['V_mean'], df['M'].map(M_to_transf_dict))]
df['V_mean'] = [c + str(v) if c != '=' else v for c, v in 
zip(df['V_mean_c'], df['V_mean'])]
df['TV_mean'] = [c + str(v) if c != '=' else v for c, v in 
zip(df['V_mean_c'], df['TV_mean'])]

结果：

# 20 s ± 289 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

@mozway的代码(M_dic根据新的M组而改变)：

def majority(s):
mode = s.mode()
return '=' if len(mode)>1 else mode.iloc[0]
M_dic = dict()
for k in M_to_transf_dict:
if M_to_transf_dict[k] == 'log10(x)':
M_dic[k] = np.log10
elif M_to_transf_dict[k] == 'exp(x)':
M_dic[k] = np.exp
df = df0.copy()
%%timeit
(df
.join(df['V']
.str.extract('(?P<V_c>D)?(?P<V_u>d+(?:.d+)?)')
.astype({'V_u': float}).fillna({'V_c': '=', 'V_u': df['V']})
)
.assign(TV=lambda d: d.groupby('M')['V_u'].apply(lambda g: M_dic[g.name](g) if g.name in M_dic else g))
.groupby(['S','M'], as_index = False)
.agg(**{'ID': ('ID', lambda x: ','.join(x.unique())),
'V_mean': ('V_u', 'mean'),
'V_mean_c': ('V_c', majority), ## FIXME
'V_N': ('V_u', 'count'),
'V_sample_sd': ('V_u', 'std'),
'TV_mean': ('TV', 'mean'),
})
.assign(TV_mean=lambda d: d['V_mean_c'].mask(d['V_mean_c'].eq('='), '')+d['TV_mean'].astype(str))
)

结果：

# 52.3 s ± 436 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

---

EDIT 3在模拟具有重复{S, M}条目的数据集后重试。

rng = default_rng(12345)
S = rng.choice(range(200000), 225000)
ID = S.copy()
S = [f"S_{Si}" for Si in S]
ID = [f"ID_{IDi}" for IDi in ID]
pM = [np.sqrt(1+i) for i in range(100)]
pM = pM / np.sum(pM)
M = rng.choice(range(100), 225000, p = pM)
M_to_transf_dict = dict()
for i in range(0,10):
M_to_transf_dict[f"M_{i}"] = 'exp(x)'
for i in range(10,30):
M_to_transf_dict[f"M_{i}"] = 'x'
for i in range(30,100):
M_to_transf_dict[f"M_{i}"] = 'log10(x)'
M = [f"M_{Mi}" for Mi in M]
S = S + S
ID = ID + ID
M = M + M
V = rng.random(450000)
Q = rng.choice(['', '<', '>'], 450000, p = [0.9, 0.05, 0.05])
V = [f"{q}{v}" for q, v in zip(Q, V)]
df0 = pd.DataFrame({'S' : S, 'ID' : ID, 'M' : M, 'V' : V})

原始代码：

10.6 s ± 154 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

@mozway的代码：

25.8 s ± 550 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

(并且输出看起来不同)。

Answer 1

由于循环、apply和eval，您的代码很慢。

这是一种更快的方法(前提是您的小组规模足够大)：

def majority(s):
mode = s.mode()
return '=' if len(mode)>1 else mode.iloc[0]
M_dic = {'Y' : np.exp, 'Z' : np.log10}
(df
.join(df['V']
.str.extract('(?P<V_c>D)?(?P<V_u>d+(?:.d+)?)')
.astype({'V_u': float}).fillna({'V_c': '=', 'V_u': df['V']})
)
.assign(TV=lambda d: d.groupby('M')['V_u'].apply(lambda g: M_dic[g.name](g) if g.name in M_dic else g))
.groupby(['S','M'], as_index = False)
.agg(**{'ID': ('ID', lambda x: ','.join(x.unique())),
'V_mean': ('V_u', 'mean'),
'V_mean_c': ('V_c', majority), ## FIXME
'V_N': ('V_u', 'count'),
'V_sample_sd': ('V_u', 'std'),
'TV_mean': ('TV', 'mean'),
})
.assign(TV_mean=lambda d: d['V_mean_c'].mask(d['V_mean_c'].eq('='), '')+d['TV_mean'].astype(str))
)

输出：

S  M       ID     V_mean V_mean_c  V_N  V_sample_sd              TV_mean
0  A  X  ID1,ID2   3.166667        =    3     2.753785   3.1666666666666665
1  B  Y      ID4  10.000000        >    1          NaN  >22026.465794806718
2  B  Z      ID4   7.000000        =    1          NaN   0.8450980400142568
3  C  Y      ID3   1.000000        <    1          NaN   <2.718281828459045