Python 基于其他列的值有条件地更改序列的值
我正在体验/学习具有以下结构的数据帧的Python:Python 基于其他列的值有条件地更改序列的值,python,pandas,dataframe,Python,Pandas,Dataframe,我正在体验/学习具有以下结构的数据帧的Python: df = pd.DataFrame({"left_color" : ["red", "green", "blue", "black", "white", ""], "right_color" : ["red", "gray", "", "black", "red", ""], "flag" : [1, 2, 3, 1, 2, 3]}) print(df
df = pd.DataFrame({"left_color" : ["red", "green", "blue", "black", "white", ""],
"right_color" : ["red", "gray", "", "black", "red", ""],
"flag" : [1, 2, 3, 1, 2, 3]})
print(df)
left_color right_color flag
0 red red 1
1 green gray 2
2 blue 3
3 black black 1
4 white red 2
5 3
我的目标是根据left_color和right_color列的值有条件地更改标志系列的值。具体而言:
如果缺少left_color或right_color,则将标志值更改为numpy NaN;
如果左侧颜色与右侧颜色不同,请将标志值更改为0。
以下是我的尝试:
def myfunc(left_side, right_side, value):
if (left_side == "") | (right_side == ""):
value = np.nan
if left_side != right_side:
value = 0
正如你所看到的,我得到的结果不是我最初描述的结果。相反,我到处都找不到价值观。以下是我想要的结果:
left_color right_color flag
0 red red 1
1 green gray 0
2 blue NaN
3 black black 1
4 white red 0
5 NaN
我想了解我的错误是什么以及如何纠正。此外,我想看看是否有一种更具python风格的方法来解决这个问题,这种方法在计算上更有效 您忘记在函数中返回值
def myfunc(left_side, right_side, value):
if (left_side == "") | (right_side == ""):
return np.nan
elif left_side != right_side:
return 0
else:
return value
df.flag=np.select([df.left_color=='',df.right_color=='', df.right_color!=df.left_color,df.right_color==df.left_color],[np.nan,np.nan,0,1] )
您忘记在函数中返回值
def myfunc(left_side, right_side, value):
if (left_side == "") | (right_side == ""):
return np.nan
elif left_side != right_side:
return 0
else:
return value
df.flag=np.select([df.left_color=='',df.right_color=='', df.right_color!=df.left_color,df.right_color==df.left_color],[np.nan,np.nan,0,1] )
您可以使用np.select,如下所示。我认为,这很可能比自定义函数快
def myfunc(left_side, right_side, value):
if (left_side == "") | (right_side == ""):
return np.nan
elif left_side != right_side:
return 0
else:
return value
df.flag=np.select([df.left_color=='',df.right_color=='', df.right_color!=df.left_color,df.right_color==df.left_color],[np.nan,np.nan,0,1] )
输出
您可以使用np.select,如下所示。我认为,这很可能比自定义函数快
def myfunc(left_side, right_side, value):
if (left_side == "") | (right_side == ""):
return np.nan
elif left_side != right_side:
return 0
else:
return value
df.flag=np.select([df.left_color=='',df.right_color=='', df.right_color!=df.left_color,df.right_color==df.left_color],[np.nan,np.nan,0,1] )
输出
您需要np.select:
输出:
left_color right_color flag
0 red red 1.0
1 green gray 0.0
2 blue NaN
3 black black 1.0
4 white red 0.0
5 NaN
您需要np.select:
输出:
left_color right_color flag
0 red red 1.0
1 green gray 0.0
2 blue NaN
3 black black 1.0
4 white red 0.0
5 NaN