根据其他两个变量创建变量

如何解决根据其他两个变量创建变量

初始问题

在Python中，我想基于c和a的值创建一个新变量b。

if a in ('GBP','AUD','CNY','NZD'):
    if b == '[00Y,01Y]':
        c= '90'
    elif b == '[01Y,02Y]':
        c = '85'
    elif b == '[02Y,03Y]':
        c = '80'
    elif b == '[03Y,04Y]':
        c = '75'
    elif b == '[04Y,05Y]':
        c = '70'
elif a in ('EUR','USD','CHF','CAD','SGD','HKD','JPY'):
    if b == '[00Y,01Y]':
        c = '95'
    elif b == '[01Y,02Y]':
        c = '90'
    elif b == '[02Y,03Y]':
        c = '85'
    elif b == '[03Y,04Y]':
        c = '80'
    elif b == '[04Y,05Y]':
        c = '75'
    elif b == '[05Y,07Y]':
        c = '60'
    elif b == '[07Y,10Y]':
        c = '55'

a和b是数据框的列，我必须使用apply才能最终获得所需的内容。

尽管这很完美，但我认为对于这么小的操作来说这是很长的代码，我想知道是否有更优雅的方法可以做到这一点。我知道np.select条件，但是它迫使我在`a上重复该条件，我发现它也不是很好。

谢谢

问题的重新形成

我最初的问题可能还不够清楚。 我想压缩以下代码而不必重复所有条件：

def f1(a,b,c,d):
    if a == 1 and b <= 5 and c in ('abc','def') and d:         s = 75
    if a == 1 and b <= 5 and c in ('abc','def') and not d:     s = 83
    if a == 1 and b <= 5 and c == 'xyz' and d:                  s = 77
    if a == 1 and b <= 5 and c == 'xyz' and not d:              s = 17
    if a == 1 and 5 < b <= 8 and c in ('abc','def') and d:     s = 28
    if a == 1 and 5 < b <= 8 and c in ('abc','def') and not d: s = 39
    if a == 1 and 5 < b <= 8 and c == 'xyz' and d:              s = 10
    if a == 1 and 5 < b <= 8 and c == 'xyz' and not d:          s = 45
    if a == 1 and b > 8 and c in ('abc','def') and d:          s = 59
    if a == 1 and b > 8 and c in ('abc','def') and not d:      s = 48
    if a == 1 and b > 8 and c == 'xyz' and d:                   s = 29
    if a == 1 and b > 8 and c == 'xyz' and not d:               s = 24
    if a == 2 and b <= 5 and c in ('abc','def') and d:         s = 39
    if a == 2 and b <= 5 and c in ('abc','def') and not d:     s = 51
    if a == 2 and b <= 5 and c == 'xyz' and d:                  s = 69
    if a == 2 and b <= 5 and c == 'xyz' and not d:              s = 42
    if a == 2 and 5 < b <= 8 and c in ('abc','def') and d:     s = 23
    if a == 2 and 5 < b <= 8 and c in ('abc','def') and not d: s = 11
    if a == 2 and 5 < b <= 8 and c == 'xyz' and d:              s = 12
    if a == 2 and 5 < b <= 8 and c == 'xyz' and not d:          s = 89
    if a == 2 and b > 8 and c in ('abc','def') and d:          s = 54
    if a == 2 and b > 8 and c in ('abc','def') and not d:      s = 23
    if a == 2 and b > 8 and c == 'xyz' and d:                   s = 22
    if a == 2 and b > 8 and c == 'xyz' and not d:               s = 98
    if a == 3 and b <= 5 and c in ('abc','def') and d:         s = 91
    if a == 3 and b <= 5 and c in ('abc','def') and not d:     s = 15
    if a == 3 and b <= 5 and c == 'xyz' and d:                  s = 55
    if a == 3 and b <= 5 and c == 'xyz' and not d:              s = 36
    if a == 3 and 5 < b <= 8 and c in ('abc','def') and d:     s = 66
    if a == 3 and 5 < b <= 8 and c in ('abc','def') and not d: s = 82
    if a == 3 and 5 < b <= 8 and c == 'xyz' and d:              s = 20
    if a == 3 and 5 < b <= 8 and c == 'xyz' and not d:          s = 98
    if a == 3 and b > 8 and c in ('abc','def') and d:          s = 77
    if a == 3 and b > 8 and c in ('abc','def') and not d:      s = 23
    if a == 3 and b > 8 and c == 'xyz' and d:                   s = 41
    if a == 3 and b > 8 and c == 'xyz' and not d:               s = 84
    return s

解决方案

我找到了使用itertools.product的解决方案。但是我们需要注意listvalues的顺序：

import numpy as np
import itertools
def f(a,d):
    listconditions = [[a==1,a==2,a==3],[b <= 5,5 < b <= 8,b > 8],[c in ("abc","def"),c == 'xyz'],[d,not d]]

        listvalues = [75,83,77,17,28,39,10,45,59,48,29,24,51,69,42,23,11,12,89,54,22,98,91,15,55,36,66,82,20,41,84]

    allcombinations = itertools.product(*listconditions)

    test = [np.logical_and.reduce(i) for i in allcombinations]

    return sum(np.array(test) * listvalues)

f(1,7,'abc',False)

39

解决方法

您可以使用字典来包含a的索引和b中的值：

options_a = {'GBP': 0,'AUD': 0,'CNY': 0,'NZD': 0,'EUR': 1,'USD': 1,'CHF': 1,'CAD': 1,'SGD': 1,'HKD': 1,'JPY': 1}

options_b = {'[00Y,01Y]': ('90','95'),'[01Y,02Y]': ('85','90'),'[02Y,03Y]': ('80','85'),'[03Y,04Y]': ('75','80'),'[04Y,05Y]': ('70','75'),'[05Y,07Y]': (None,'60'),'[07Y,10Y]': (None,'55')}

# Get the index of the tuple by looking up 'a' first
idx = options_a[a]

# Then use that index when you look up 'b' to grab the correct value for 'c'
c = options_b[b][idx]

如果您获得了您不打算使用的任何组合，则会产生一个KeyError，您可能会或可能不想处理：

try:
    idx = options_a[a]
    tup = options_b[b]
except KeyError:
    print("Do something")
else:
    c = tup[idx]

,

使用for循环和列表做您想要的事情，我假设您需要将值减少5

tupel1 =  ('GBP','AUD','CNY','NZD')
tuple2 = ('EUR','USD','CHF','CAD','SGD','HKD','JPY')
listb = ['[00Y,01Y]',03Y]',04Y]',05Y]',]

for i in range(listb):
  if listb[i]==b:
    if a in tuble1:
       c = str(90 - 5*i)
    elif a in tuble2:
       c = str(95 -5*i)
  
 

,

另一种选择，单线：

(95 if a in {'EUR','JPY'} else 90) - 5 * [0,1,2,3,4,5,7].index(int(b[1:3]))

或者将其包装得更多，但会变脏：

(95 if a in 'EUR USD CHF CAD SGD HKD JPY' else 90) - 5 * [0,7].index(int(b[1:3]))