numpy vectorize np.prod 无法构造超过 32 个操作数的 ufunc

Question

我知道这里有一个类似的问题：Python numpy.vectorize: ValueError: Cannot construct a ufunc with more than 32 operands

但我的情况不同。

我有一个 32 列的 df，你可以通过运行以下代码来获得它：

import numpy as np
import pandas as pd
from io import StringIO
dfs = """
    M0  M1  M2  M3 M4  M5 M6 M7 M8 M9 M10 M11 M12 M13 M14 M15 M16 M17 M18 M19 M20 M21 M22 M23 M24 M25 M26 M27 M28 M29 M30  age 
1   1   2   3    4  5   6  1  2 3    4  5  6   1   2    3  4  5    6   7   8    9 1    2  3    4  5    6  1    2   3    4   3.2        
2   7   5   4    5  8   3  1  2 3    4  5  6   1   2    3  4  5    6   7   8    9 1    2  3    4  5    6  1    2   3    4   4.5
3   4   8   9    3  5   2  1  2 3    4  5  6   1   2    3  4  5    6   7   8    9 1    2  3    4  5    6  1    2   3    4   6.7
"""
df = pd.read_csv(StringIO(dfs.strip()), sep='\s+', )
df

基于业务逻辑我构建了一个向量化的函数，如果函数的参数总数小于32就可以了：

M=["M0","M1","M2","M3","M4","M5","M6","M7","M8","M9","M10","M11","M12","M13","M14","M15","M16","M17","M18","M19",
       "M20","M21","M22","M23","M24","M25","M26","M27","M28","M29"]
    
    def func2(df, M):
        return [df[i].values for i in M] 
    
    def func(age,*Ms):
        newcol=np.prod(Ms[0:age])
        return newcol
    
    vfunc = np.frompyfunc(func, len(M)+1, 1)
    
    df['newcol']=vfunc(df['age'].values.astype(int), *func2(df,M))

为了便于理解，func2只是让代码更简洁，它为func生成所有参数，没有func2的代码如下：

def func(age,M0,M1,M2,...,M29):
    newcol=np.prod(Ms[0:age])
    return newcol

vfunc = np.frompyfunc(func, 31, 1)

df['newcol']=vfunc(df['age'].values.astype(int), df['M1'].values,...,df['M29'].values)

真正的问题是，一旦参数的数量等于或大于 32，如下所示：

M=["M0","M1","M2","M3","M4","M5","M6","M7","M8","M9","M10","M11","M12","M13","M14","M15","M16","M17","M18","M19",
           "M20","M21","M22","M23","M24","M25","M26","M27","M28","M29","M30"] # M30 is the only difference from the above function
        
        def func2(df, M):
            return [df[i].values for i in M] 
        
        def func(age,*Ms):
            newcol=np.prod(Ms[0:age])
            return newcol
        
        vfunc = np.frompyfunc(func, len(M)+1, 1)
        
        df['newcol']=vfunc(df['age'].values.astype(int), *func2(df,M))

我收到错误：

ValueError                                Traceback (most recent call last)
<ipython-input-66-9a042ad44f9b> in <module>()
     76     return newcol
     77 
---> 78 vfunc = np.frompyfunc(func, len(M)+1, 1)
     79 
     80 df['newcol']=vfunc(df['age'].values.astype(int), *func2(df,M))

ValueError: Cannot construct a ufunc with more than 32 operands (requested number were: inputs = 32 and outputs = 1)

在我的真实业务逻辑中，我有超过 100 列需要使用 np.pro 来计算，所以这真的让我很困惑。有朋友可以帮忙吗？

Answer 1

这是实现您的结果的一种方法。选中所有带有filter的M列，使用where将列位置高于年龄列的所有值替换为nan，然后沿列使用prod。

df['newcol'] = (
     # keep only Mx columns
    df.filter(like='M')
      # keep only the values when the position of the column
      # is less than the age
      .where(lambda x: (np.arange(x.shape[1])+1)<df['age'].to_numpy()[:, None])
      # multiply all the non-nan values per row
      .prod(axis=1)
)
print(df)