在其他列之间创建具有各种条件逻辑的新列

Question

我有以下数据集

test = pd.DataFrame({'date':['2018-08-01','2018-08-02','2018-08-03','2019-09-01','2019-09-02','2019-09-03','2020-01-02','2020-01-03','2020-01-04','2020-10-04','2020-10-05'],
                    'account':['a','a','a','b','b','b','c','c','c','d','e'],
                    'tot_chg':[2072,2072,2072,322,322,322,483,483,483,140,570],
                    'denied':[1878,1036,1036,322,161,161,150,322,322,105,570],
                    'denied_sum':[1878,2914,3950,322,483,644,150,472,794,105,570]})

我想在 append 中创建一个基于以下参数的名为denied_true的新列：

1. 当denied_sum小于tot_chgs时，返回denied
2. 直到denied_sum超过tot_chgs ，然后计算所有先前denied_true的总和减去tot_chgs之间的剩余差异
3. 如果denied曾经等于tot_chgs在第一个实例中，则返回denied并使帐户的剩余行为 0

output 实际上应该是这样的：

output 的 dataframe 是：

output = pd.DataFrame({'date':['2018-08-01','2018-08-02','2018-08-03','2019-09-01','2019-09-02','2019-09-03','2020-01-02','2020-01-03','2020-01-04','2020-10-04','2020-10-05'],
                    'account':['a','a','a','b','b','b','c','c','c','d','e'],
                    'tot_chg':[2072,2072,2072,322,322,322,483,483,483,140,570],
                    'denied':[1878,1036,1036,322,161,161,150,322,322,105,570],
                    'denied_sum':[1878,2914,3950,322,483,644,150,472,794,105,570],
                    'denied_true':[1878,194,0,322,0,0,150,322,11,105,570]})

到目前为止，我已经使用 where 尝试了以下代码，但它缺少从 tot_chgs 中减去先前的 denied_true 值的条件

test['denied_true'] = test.denied_sum.to_numpy()
test.denied_true.where(test.denied_sum.le(test.tot_chg),other=0,inplace=True)
test

但是，我不太确定如何将 append 多个条件设置为 function。也许我需要 if/elif 循环，或者 boolean 掩码。 任何帮助将不胜感激！

Answer 1

您可以将 DataFrame 转换为 OrderedDict 并以这种直接的方式处理它：

import pandas as pd
from collections import OrderedDict

test = pd.DataFrame({'date':      ['2018-08-01', '2018-08-02', '2018-08-03', '2019-09-01', '2019-09-02', '2019-09-03', '2020-01-02', '2020-01-03', '2020-01-04', '2020-10-04', '2020-10-05'],
                    'account':    ['a', 'a', 'a', 'b', 'b', 'b', 'c', 'c', 'c', 'd', 'e'],
                    'tot_chg':    [2072, 2072, 2072, 322, 322, 322, 483, 483, 483, 140, 570],
                    'denied':     [1878, 1036, 1036, 322, 161, 161, 150, 322, 322, 105, 570],
                    'denied_sum': [1878, 2914, 3950, 322, 483, 644, 150, 472, 794, 105, 570]})
        
# convert DataFrame into OrderedDict
od = test.to_dict(into=OrderedDict)

# functions (samples)
def zero(dict, row):
    # if denied == denied_sum
    # change the dict...
    return dict['denied'][row]

def ex(dict, row):
    # if exceeds
    # change the dict...
    return 'exceed()'

def eq(dict, row):
    # if equals
    # change the dict...
    return 'equal()'

def get_value(dict, row):
    # conditions
    if dict['denied'][row]     == dict['denied_sum'][row]: return zero(dict, row)
    if dict['denied_sum'][row] <  dict['tot_chg'][row]:    return dict['denied'][row]
    if dict['denied_sum'][row] >  dict['tot_chg'][row]:    return ex(dict, row)
    if dict['denied_sum'][row] == dict['tot_chg'][row]:    return eq(dict, row)


# MAIN

# make a list (column) of 'denied_true' values
denied_true_list = [(row, get_value(od, row)) for row in range(len(od["date"]))]

# convert the list into a dict
denied_true_dict = {'denied_true': OrderedDict(denied_true_list)}

# add the dict to the OrderedDict
od.update(OrderedDict(denied_true_dict))

# convert the OrderedDict back into DataFrame
test = pd.DataFrame(od)

输入：

          date account  tot_chg  denied  denied_sum
0   2018-08-01       a     2072    1878        1878
1   2018-08-02       a     2072    1036        2914
2   2018-08-03       a     2072    1036        3950
3   2019-09-01       b      322     322         322
4   2019-09-02       b      322     161         483
5   2019-09-03       b      322     161         644
6   2020-01-02       c      483     150         150
7   2020-01-03       c      483     322         472
8   2020-01-04       c      483     322         794
9   2020-10-04       d      140     105         105
10  2020-10-05       e      570     570         570

Output：

          date account  tot_chg  denied  denied_sum denied_true
0   2018-08-01       a     2072    1878        1878        1878
1   2018-08-02       a     2072    1036        2914    exceed()
2   2018-08-03       a     2072    1036        3950    exceed()
3   2019-09-01       b      322     322         322         322
4   2019-09-02       b      322     161         483    exceed()
5   2019-09-03       b      322     161         644    exceed()
6   2020-01-02       c      483     150         150         150
7   2020-01-03       c      483     322         472         322
8   2020-01-04       c      483     322         794    exceed()
9   2020-10-04       d      140     105         105         105
10  2020-10-05       e      570     570         570         570

我没有在函数中完全实现你的逻辑，因为它只是一个示例。

大致相同（可能会更容易一些）可以通过 DataFrame > JSON > DataFrame 完成。

---

更新。 我试图实现 function ex() 。 这是它的样子。

def ex(dict, row):
    # if exceeds
    denied_true_slice = denied_true_list[0:row] # <-- global list
    tot_chg_slice     = [dict['tot_chg'][r] for r in range(row)]
    denied_true_sum   = sum ([v for r, v in enumerate(denied_true_slice) if tot_chg_slice[r] > v])
    value = tot_chg_slice[-1] - denied_true_sum
    return value if value > 0 else 0

我不太确定它是否按预期工作。 由于我不完全了解古怪的条件。 但我确信它看起来相当丑陋和神秘，并且可能与最佳 Stackoverflow 的示例不一致。

现在有了全局列表，所以，MAIN 部分现在看起来像这样：

# MAIN

# make a list (column) of 'denied_true' values
denied_true_list = [] # <-- the global list
for row, _ in enumerate(od['date']):
    denied_true_list.append(get_value(od,row))

denied_true_list = [(row, value) for row, value in enumerate(denied_true_list)]

# convert the list into a dict
denied_true_dict = {'denied_true': OrderedDict(denied_true_list)}

# add the dict to the OrderedDict
od.update(OrderedDict(denied_true_dict))

# convert the OrderedDict back into DataFrame
test = pd.DataFrame(od)

Output：

          date account  tot_chg  denied  denied_sum  denied_true
0   2018-08-01       a     2072    1878        1878         1878
1   2018-08-02       a     2072    1036        2914          194
2   2018-08-03       a     2072    1036        3950            0
3   2019-09-01       b      322     322         322          322
4   2019-09-02       b      322     161         483            0
5   2019-09-03       b      322     161         644            0
6   2020-01-02       c      483     150         150          150
7   2020-01-03       c      483     322         472          322
8   2020-01-04       c      483     322         794            0
9   2020-10-04       d      140     105         105          105
10  2020-10-05       e      570     570         570          570

我相信通过本机 Pandas 工具可以做得更漂亮。