【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"]
【问题讨论】: