pandas 'DataFrame' 对象没有属性 'map'
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pandas 'DataFrame' object has no attribute 'map'
提问by daiyue
I have two df - df_a and df_b,
我有两个 df - df_a 和 df_b，
# df_a
number cur code
1000 USD 700
2000 USD 800
3000 USD 900
# df_b
number amount deletion code
1000 0.0 L 700
1000 10.0 X 700
1000 10.0 X 700
2000 20.0 X 800
2000 20.0 X 800
3000 0.0 L 900
3000 0.0 L 900
I want to left merge df_awith df_b,
我想左合并df_a有df_b，
df_a = df_a.merge(df_b.loc[df_b.deletion != 'L'], how='left', on=['number', 'code'])
and also, create a flag called deletedin the merge result df_a, that has three possible values - full, partial and none;
并且，创建一个deleted在合并结果中调用的标志，该标志df_a具有三个可能的值 - full、partial 和 none；
full- if all rows associated with a particular numbervalue, have deletion= L;
full- 如果所有行都与特定number值相关联，则有deletion= L；
partial- if some rows associated with a particular numbervalue, have deletion= L;
partial- 如果某些行与特定number值相关联，则有deletion= L；
none- no rows associated with a particular numbervalue, have deletion= L;
none- 没有与特定number值相关联的行，有deletion= L；
Also when doing the merge, rows from df_bwith deletion= L should not be considered; so the result looks like,
此外，在进行合并时，不应考虑来自df_bwith deletion= L 的行；所以结果看起来像，
number amount deletion deleted cur code
1000 10.0 X partial USD 700
1000 10.0 X partial USD 700
2000 20.0 X none USD 800
2000 20.0 X none USD 800
3000 0.0 NaN full USD 900
I tried,
我试过，
g = df_b['deletion'].ne('L').groupby([df_b['number'], df_b['code']])
m1 = g.any()
m2 = g.all()
d1 = dict.fromkeys(m1.index[m1 & ~m2], 'partial')
d2 = dict.fromkeys(m2.index[m2], 'full')
d = {**d1, **d2}
df_a = df_a.merge(df_b.loc[df_b.deletion != 'L'], how='left', on=['code', 'number'])
df_a['deleted'] = df_a[['number', 'code']].map(d).fillna('none')
but I got an error,
但我有一个错误，
AttributeError: 'DataFrame' object has no attribute 'map'
It seems dfdoes not have mapfunction, so I am wondering if there are any alternative ways to achieve this.
它似乎df没有map功能，所以我想知道是否有其他方法可以实现这一点。
回答by jpp
pd.DataFrameobjects don't have a mapmethod. You can instead construct an index from two columns and use pd.Index.mapwith a function:
pd.DataFrame对象没有map方法。您可以改为从两列构造索引并pd.Index.map与函数一起使用：
df_a['deleted'] = df_a.set_index(['number', 'code']).index.map(d.get)
df_a['deleted'] = df_a['deleted'].fillna('none')
Notice we use d.getinstead of dhere. Unlike pd.Series.map, pd.Index.mapcannot accept a dictionary directly, but it can accept a function such as dict.get.
注意我们d.get在d这里使用了代替。与 不同pd.Series.map，pd.Index.map不能直接接受字典，但可以接受诸如dict.get.
Note also we split apart the fillnaoperation as pd.Index.mapreturns an array rather than a series.
另请注意，我们将fillna操作拆分为pd.Index.map返回一个数组而不是一个系列。
