根据多列条件返回值

Posted 2023-03-11

【中文标题】根据多列条件返回值

【英文标题】：Return value based on condition from multiple columns

【发布时间】：2020-10-23 08:45:26

【问题描述】：

我有一个包含 4 列的数据框，基本上我正在尝试使用 if 语句创建另一列并返回满足条件的值

If NR/HL1 is not equal to 0, then outputColumn(NR/HL) = NR/HL1

if NR/HL1 is equals to 0, then outputColumn(NR/HL) = NR/HL2

if NR/HL1 is equals to 0 and NR/HL2 is equal to 0, then outputColumn(NR/HL) = NR/HL3

SKU NR/HL1 NR/HL2 NR/HL3  OutputColumn(NR/HL)
123  10     20     0         10
456  0      30     20        30
567  0      0      40        40
890  10     20     50        10

我使用了下面的代码，它工作正常，但不是 100% 准确。总是错过一个或另一个条件。如果您检查输出条件 3 的图像是否满足但它返回默认值。 NR/HL3 !=0, But still NR/HL ==0

def f(AC_off_trade):
    if AC_off_trade['NR/HL1'] != 0:
        return AC_off_trade['NR/HL1']
    if AC_off_trade['NR/HL1'] == 0:
        val = AC_off_trade['NR/HL2']
    if AC_off_trade['NR/HL1'] == 0 and AC_off_trade['NR/HL2'] == 0:
        return AC_off_trade['NR/HL3']
    else:
        return 0
AC_off_trade['NR/HL'] = AC_off_trade.apply(f,axis=1)

更新代码

#defining condition
hl1_equal_0_condition = AC_off_trade["NR/HL1"]==0.0    
hl2_equal_0_contition = AC_off_trade["NR/HL2"]==0.0
#default value
AC_off_trade.loc[:,"NR/HL"]=0
#setting values depending on condition
AC_off_trade.loc[~hl1_equal_0_condition, "NR/HL"] = AC_off_trade["NR/HL1"]
AC_off_trade.loc[hl1_equal_0_condition, "NR/HL"] = AC_off_trade["NR/HL2"]
AC_off_trade.loc[hl1_equal_0_condition & hl2_equal_0_contition, "NR/HL"] = AC_off_trade["NR/HL3"]

【参考方案1】：

首先，您应该尽量避免使用apply 来支持矢量化，因为性能要好得多。然后你可以使用条件来获得你想要的输出：

df = pd.DataFrame("SKU": [123, 456, 567, 890], "NR/HL1" : [10, 0, 0, 10], "NR/HL2": [20, 30, 0, 20],"NR/HL3": [0, 20, 40, 50])

# Defining conditions 
hl1_equal_0_condition = df["NR/HL1"]==0    
hl2_equal_0_contition = df["NR/HL2"]==0

# Setting the default value
df.loc[:,"NR/HL"] = 0

# Setting the values deppeding on tje conditions
df.loc[~hl1_equal_0_condition, "NR/HL"] = df["NR/HL1"]
df.loc[hl1_equal_0_condition, "NR/HL"] = df["NR/HL2"]
df.loc[hl1_equal_0_condition & hl2_equal_0_contition, "NR/HL"] = df["NR/HL3"]

输出：

    SKU NR/HL1  NR/HL2  NR/HL3  NR/HL
0   123 10      20      0       10
1   456 0       30      20      30
2   567 0       0       40      40
3   890 10      20      50      10

【讨论】：

感谢您的提示和代码，但它仍然面临同样的问题。条件 3 不起作用。 NR/HL1 and NR/HL2 = 0 but NR/HL3 != 0 but NR/HL return 0。我已将屏幕截图图像附加到问题中

条件 3 对我有用，应该还有别的东西。你也可以给我看代码吗？您是否更改了条件的顺序？你应该保持原样。另一件事，您的列类型（NR/HL2 和 NR/HL1）可能是一个字符串并且在比较时会失败？

成功了，一列中有字符串。我将str 更改为```int`。再次感谢您的帮助