熊猫数据框比较和替换值

时间:2020-06-09 11:47:07

标签: python python-3.x pandas

我有两个如下所示的熊猫数据框。 “否”列是一个公共字段。基于“否”,我想替换第一个数据框列“总计”中的值。

条件为:如果“否”匹配,则从dataframe2获取“ Marks1”值,然后在“总计”列中进行替换。如果'Marks1'为NULL,则获取'Marks2'值并替换为'Total'。如果两个(Marks1 / Marks2)都为空,请在“总计”列中将其替换为空。 最终结果应该在数据帧1中。这两个数据帧都有几十万条记录。

Data frame1
No|Total
1234|11
2515|21
3412|32
4854|
7732|53

Data frame2
No|Marks1|Marks2
1234|99|23
2515|98|31
3412||20
4854||98
7732||

Result :
No|Total
1234|99
2515|98
3412|20
4854|98
7732|

2 个答案:

答案 0 :(得分:2)

使用Series.map并用Series.fillnaMarks1替换缺失的值Marks2

df = df2.set_index('No')

df1['Total'] = df1['No'].map(df['Marks1'].fillna(df['Marks2']))
print (df1)
     No  Total
0  1234   99.0
1  2515   98.0
2  3412   20.0
3  4854   98.0
4  7732    NaN

如果Nodf2中可能有重复的值,请使用:

print (df2)
     No  Marks1  Marks2
0  1234    99.0    23.0 <- duplicated No
1  1234    98.0    31.0 <- duplicated No
2  3412     NaN    20.0
3  4854     NaN    98.0
4  7732     NaN     NaN

#newer pandas versions
df = df2.set_index('No').sum(level=0, min_count=1)
#oldier pandas versions
#df = df2.set_index('No').sum(level=0)
print (df)
      Marks1  Marks2
No                  
1234   197.0    54.0<- unique No, values are summed per index created by No
3412     NaN    20.0
4854     NaN    98.0
7732     NaN     NaN

df1['Total'] = df1['No'].map(df['Marks1'].fillna(df['Marks2']))
print (df1)
     No  Total
0  1234  197.0
1  2515    NaN
2  3412   20.0
3  4854   98.0
4  7732    NaN

如果df1df2中的索引值相同,并且每个No值匹配,则使用:

df1['Total'] = df2['Marks1'].fillna(df2['Marks2'])

答案 1 :(得分:1)

您可以在此处使用np.select

m = df2['Marks1'].notna()
m1 = df2['Marks1'].isna() & df2['Marks2'].notna()
condlist = [m,m1]
choice = [df2['Marks1'] , df2['Marks2']]
df1['Total'] = np.select(condlist,choice,np.nan)

     No  Total
0  1234   99.0
1  2515   98.0
2  3412   20.0
3  4854   98.0
4  7732    NaN