如何在 Python DataFrame 上执行 itertools 的置换操作？

Question

我在 Python 中有以下数据框 test ：

test = pd.DataFrame({'id':['F8510004123','A3100002543','Z3510002123'],
                     'product':['retail','retail','others'],
                     'type':['E','E','D'],
                     'quantity':[25,34,150],
                     'nro_ope':[2,3,26],
                     'payments':[[1030.97,1030.97,584.91],[1610.74,1610.74,1610.74,1610.74,1611.14],[1007.52,1007.52,1007.52,1007.52,500,500,852.95]]
                     })
test.dtypes

用表格表示：

id	product	types	quantity	nro_ope	payments
F8510004123	retail	E	25	2	1030.97,1030.97,584.91
A3100002543	retail	E	34	3	1610.74,1610.74,1610.74,1610.74,1611.14
Z3510002123	others	D	150	26	1007.52,1007.52,1007.52,1007.52,500,500,852.95

根据数据类型：

test.dtypes

输出[449]：

id 对象

产品对象

类型对象

数量 int64

支付对象

dtype: 对象

我想使用库 itertools 对数据框 test 的 payments 列的每个元素执行以下置换操作Python：

  payment_column_1 = [1030.97,1030.97,584.91]
  payment_column_2 = [1610.74,1610.74,1610.74,1610.74,1611.14]
  payment_column_3 = [1007.52,1007.52,1007.52,1007.52,500,500,852.95]
 
 
  from itertools import permutations
  permuted_operation_1 = [round(abs(a/b -1),3) for a,b in permutations(payment_column_1,2)]
  permuted_operation_2 = [round(abs(a/b -1),3) for a,b in permutations(payment_column_2,2)]
  permuted_operation_3 = [round(abs(a/b -1),3) for a,b in permutations(payment_column_3,2)]

输出：

print(permuted_operation_1)

[0.0, 0.763, 0.0, 0.763, 0.433, 0.433]

print(permuted_operation_2)

[0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0]

print(permuted_operation_3)

[0.0, 0.0, 0.0, 1.015, 1.015, 0.181, 0.0, 0.0, 0.0, 1.015, 1.015, 0.181, 0.0, 0.0, 0.0, 1.015, 1.015, 0.181, 0.0, 0.0, 0.0, 1.015, 1.015, 0.181, 0.504, 0.504, 0.504, 0.504, 0.0, 0.414、0.504、0.504、0.504、0.504、0.0、0.414、0.153、0.153、0.153、0.153、0.706、0.706]

但是，如您所见，当我需要来自同一数据框 test 创建一个新列 var_payments 时，我从列表中应用此置换操作插入数组，其中包含数据框具有的每一行的操作结果，如下表所示：

id	product	types	quantity	nro_ope	payments	var_payments
F8510004123	retail	E	25	2	{1030.97,1030.97,584.91}	{0.0, 0.763, 0.0, 0.763, 0.433, 0.433}
A3100002543	retail	E	34	3	{1610.74,1610.74,1610.74,1610.74,1611.14}	{0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0}
Z3510002123	others	D	150	26	{1007.52,1007.52,1007.52,1007.52,500,500,852.95}	{0.0, 0.0, 0.0, 1.015, 1.015, 0.181, 0.0, 0.0, 0.0, 1.015, 1.015, 0.181, 0.0, 0.0, 0.0, 1.015, 1.015, 0.181, 0.0, 0.0, 0.0, 1.015, 1.015, 0.181, 0.504, 0.504, 0.504, 0.504, 0.0, 0.414, 0.504, 0.504, 0.504, 0.504, 0.0, 0.414, 0.153, 0.153, 0.153, 0.153, 0.706, 0.706}

我在以下方面取得了进展，但显然是错误的：

test['var_payments'] = test['payments'].apply( lambda r:round(abs(a/b -1),3) for a,b in 
permutations (r,2))

输出：

NameError: name 'r' 未定义

最后，我需要创建列flag，它基本上遍历了新创建的列var_payments的元素，如果发现至少有两个元素大于或等于0.05，则为Yes，否则为No

id	product	types	quantity	nro_ope	payments	var_payments	flag
F8510004123	retail	E	25	2	{1030.97,1030.97,584.91}	{0.0, 0.763, 0.0, 0.763, 0.433, 0.433}	Yes
A3100002543	retail	E	34	3	{1610.74,1610.74,1610.74,1610.74,1611.14}	{0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0}	No
Z3510002123	others	D	150	26	{1007.52,1007.52,1007.52,1007.52,500,500,852.95}	{0.0, 0.0, 0.0, 1.015, 1.015, 0.181, 0.0, 0.0, 0.0, 1.015, 1.015, 0.181, 0.0, 0.0, 0.0, 1.015, 1.015, 0.181, 0.0, 0.0, 0.0, 1.015, 1.015, 0.181, 0.504, 0.504, 0.504, 0.504, 0.0, 0.414, 0.504, 0.504, 0.504, 0.504, 0.0, 0.414, 0.153, 0.153, 0.153, 0.153, 0.706, 0.706}	Yes

我该怎么做？

我很关注你们的cmets。

感谢支持。

Answer 1

我是通过在 python 中创建函数得到的：

test= pd.DataFrame({'id':['F8510004123','A3100002543','Z3510002123'],
                     'product':['retail','retail','others'],
                     'type':['E','E','D'],
                     'quantity':[25,34,150],
                     'nro_ope':[2,3,26],
                     'payments':[[1030.97,1030.97,584.91],[1610.74,1610.74,1610.74,1610.74,1611.14],[1007.52,1007.52,1007.52,1007.52,500,500,852.95]]
                     })

;

def var_payments (x) :
    variation  = [round(abs(a/b -1),3) for a,b in permutations(x,2)]
    return variation 

;

def count_var_pay (x) :
    count = 0
    for element in x:
        if element >= 0.05 :
            count += 1
        else:
             0                    
    return count

;

def flag_var_payments (x) : 
    if x >= 2 :
        return 'Yes'
    else : 
        return 'No'

;

test['var_payments'] = test.payments.apply(lambda x:var_payments(x))

test['count_p'] = test.var_payments.apply(lambda x:count_var_pay(x))

test['flag'] = test.count_p.apply(lambda x:flag_var_payments(x))