給定一個數值數據框，我想對所有列組合執行加、減、乘和除。對于 3 及以上的組合，最快的方法是什么？下面給出了一個最小的可重現示例，其中包含 2。import numpy as npimport pandas as pdfrom itertools import combinationsfrom itertools import permutationsfrom sklearn.datasets import load_boston # the datasetX, y = load_boston(return_X_y=True)X = pd.DataFrame(X)combos2 = list(combinations(X.columns,2))perm3 = list(permutations(X.columns,3)) # how would i do this with out typing out all the permutationsfor i in combos2: X[f'{i[0]}_X_{i[1]}'] = X.iloc[:,i[0]]*X.iloc[:,i[1]] # Multiply X[f'{i[0]}_+_{i[1]}'] = X.iloc[:,i[0]]+X.iloc[:,i[1]] # Add X[f'{i[0]}_-_{i[1]}'] = X.iloc[:,i[0]]-X.iloc[:,i[1]] # Subtract X[f'{i[0]}_/_{i[1]}'] = X.iloc[:,i[0]]/(X.iloc[:,i[1]]+1e-20) # Divide我正在考慮一種將“運算符 + * - / 添加到組合中的方法，這樣它可以用比手動輸入所有組合更少的行來編寫，但我不知道從哪里開始？我想要所有訂單：即 (a * b + c) , (a * b - c) , (a * b / c) 等理想情況下不要留下重復的列。即（a + b + c）和（c + b + a）例如，如果我有 3 列 ab c。我想要一個新列 (a * b + c)。
2 回答
蝴蝶不菲
TA貢獻1810條經驗 獲得超4個贊
我希望這會幫助您入門：
operators = ['-', '+', '*', '/']
operands = ['a', 'b', 'c']
# find out all possible combination of operators first. So if you have 3 operands, that would be all permutations of the operators, taken 2 at a time. Also append the same expression operator combinations to the list
from itertools import permutations
operator_combinations = list(permutations(operators, len(operands)-1))
operator_combinations.extend([op]*(len(operands)-1) for op in operators)
# create a list for each possible expression, appending it with an operand and then an operator and so on, finishing off with an operand.
exp = []
for symbols in operator_combinations:
temp = []
for o,s in zip(operands, symbols):
temp.extend([o,s])
temp.append(operands[-1])
exp.append(temp)
for ans in exp:
print(''.join(ans))
輸出 ：
a-b+c
a-b*c
a-b/c
a+b-c
a+b*c
a+b/c
a*b-c
a*b+c
a*b/c
a/b-c
a/b+c
a/b*c
a-b-c
a+b+c
a*b*c
a/b/c
慕慕森
TA貢獻1856條經驗 獲得超17個贊
這是一個天真的解決方案，它輸出所有列的 2 和 3 的組合。
組合列表
使用 operator 包創建一個函數
for循環組合
這可能有重復的列，因此刪除重復項
from sklearn.datasets import load_boston
from itertools import combinations
import operator as op
X, y = load_boston(return_X_y=True)
X = pd.DataFrame(X)
comb= list(combinations(X.columns,3))
def operations(x,a,b):
if (x == '+'):
d = op.add(a,b)
if (x == '-'):
d = op.sub(a,b)
if (x == '*'):
d = op.mul(a,b)
if (x == '/'): # divide by 0 error
d = op.truediv(a,(b + 1e-20))
return d
for x in ['*','/','+','-']:
for y in ['*','/','+','-']:
for i in comb:
a = X.iloc[:,i[0]].values
b = X.iloc[:,i[1]].values
c = X.iloc[:,i[2]].values
d = operations(x,a,b)
e = operations(y,d,c)
X[f'{i[0]}{x}{i[1]}{y}{i[2]}'] = e
X[f'{i[0]}{x}{i[1]}'] = d
X = X.loc[:,~X.columns.duplicated()]
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