【问题标题】:Pandas Data Frame conditional flow with multiple columns具有多列的 Pandas Dataframe 条件流
【发布时间】:2019-02-14 13:05:57
【问题描述】:

我有一个如下的数据框:

fix = pd.DataFrame()
fix ['Home'] =['A','B','C','D','E']
fix ['Away'] =['F','G','H','I','J']
fix ['GD = -2'] = [0.2,0.3,0.5,0.1,0.6]
fix ['GD = -1'] = [0.25,0.1,0.55,0.35,0.43]
fix ['GD = 0'] = [0.1,0.2,0.23,0.5,0.4]
fix ['GD = 2'] = [0.1,0.5,0.2,0.12,0.18]
fix ['GD = 1'] = [0.24,0.5,0.33,0.31,0.13]

我想创建一个新列,其中包含基于 GD 的获胜球队(即 GD +ve 表示主队获胜,GD -Ve 表示客队获胜,GD = 0 表示平局。

所以我编写了以下代码来锻炼新列。

GDPlus = fix ['GD=1'] or fix['GD=2']
GDMins = fix ['GD= -1'] or fix['GD= -2'] 
fix['Winning_Team'] = np.select([GDPlus,GDMins],[fix.Home,fix.Away],default ='Draw')

它给我一个错误如下:

ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().

谁能告诉我该怎么做?

【问题讨论】:

  • 在 pandas 中按位 OR 使用 |,但它用于链接布尔掩码。
  • 预期输出是什么?
  • 这些是赔率吗?赔率最高的列应该是结果?

标签: python pandas dataframe conditional-operator


【解决方案1】:

如果想要max 值的新列:

#get max values per rows
smax = fix.max(axis=1)

compare by eq (==) and check if at least one True per rows
GDPlus = fix[['GD = 1','GD = 2']].eq(smax, axis=0).any(axis=1)
GDMins = fix[['GD = -1','GD = -2']].eq(smax, axis=0).any(axis=1)

您的解决方案应该通过eq (==) 进行比较来更改:

GDPlus = fix ['GD = 1'].eq(smax) | fix['GD = 2'].eq(smax)
GDMins = fix ['GD = -1'].eq(smax) | fix['GD = -2'] .eq(smax)

#alternative solution
#GDPlus = (fix['GD = 1'] == smax) | (fix['GD = 2'] == smax)
#GDMins = (fix['GD = -1'] == smax) | (fix['GD = -2'] == smax)

fix['Winning_Team'] = np.select([GDPlus,GDMins],[fix.Home,fix.Away],default ='Draw')
print (fix)
  Home Away  GD = -2  GD = -1  GD = 0  GD = 2  GD = 1 Winning_Team
0    A    F      0.2     0.25    0.10    0.10    0.24            F
1    B    G      0.3     0.10    0.20    0.50    0.50            B
2    C    H      0.5     0.55    0.23    0.20    0.33            H
3    D    I      0.1     0.35    0.50    0.12    0.31         Draw
4    E    J      0.6     0.43    0.40    0.18    0.13            J

【讨论】:

  • 谢谢。杰兹。有用。您能否解释一下以下两行:GDPplus = fix ['GD = 1'].eq(smax) | fix['GD = 2'].eq(smax) GDMins = fix ['GD = -1'].eq(smax) |修复['GD = -2'] .eq(smax)
  • .eq 用作 =?
  • 亲爱的 Jez,感谢您更新解决方案
猜你喜欢
  • 2016-10-06
  • 1970-01-01
  • 1970-01-01
  • 2017-07-25
  • 2016-07-02
  • 2017-03-01
  • 2018-12-17
  • 2018-05-05
  • 1970-01-01
相关资源
最近更新 更多