Question

我有两个数据框，如下所示：

df1：

Cell    NodeName        conc        Delta
S1C1    B4MU1241    B4MU1241;S1C1   0.2
S2C1    B4MU1241    B4MU1241;S2C1   0.2
S3C1    B4MU1241    B4MU1241;S3C1   1
S4C1    B4MU1241    B4MU1241;S4C1   11.1
S1C1    B4MU1702    B4MU1702;S1C1   0.2
S1C2    B4MU1702    B4MU1702;S1C2   0.2
S2C1    B4MU1702    B4MU1702;S2C1   0.1
S2C2    B4MU1702    B4MU1702;S2C2   0
S3C1    B4MU1702    B4MU1702;S3C1   0.1
S3C2    B4MU1702    B4MU1702;S3C2   0.2
S4C1    B4MU1702    B4MU1702;S4C1   0.1
S4C2    B4MU1702    B4MU1702;S4C2   0.1

df2：

Cell        NodeName      conc       Temparature-DUW    Delta
S1C1;       B4MU1241    B4MU1241;S1C1       60C 
S2C1;       B4MU1241    B4MU1241;S2C1       60C 
S3C1;       B4MU1241    B4MU1241;S3C1       60C 
S4C1;       B4MU1241    B4MU1241;S4C1       60C 
S1C1;S1C2;  B4MU1702    B4MU1702;S1C1;S1C2  56C 
S2C1;S2C2;  B4MU1702    B4MU1702;S2C1;S2C2  56C 
S3C1;S3C2;  B4MU1702    B4MU1702;S3C1;S3C2  56C 
S4C1;S4C2;  B4MU1702    B4MU1702;S4C1;S4C2  56C

现在，我要在df2中填充列“ Delta”，以使输出应为：

Cell            NodeName    conc        Temparature-DUW Delta
S1C1;       B4MU1241    B4MU1241;S1C1       60C          0.2
S2C1;       B4MU1241    B4MU1241;S2C1       60C          0.2
S3C1;       B4MU1241    B4MU1241;S3C1       60C           1
S4C1;       B4MU1241    B4MU1241;S4C1       60C          11.1
S1C1;S1C2;  B4MU1702    B4MU1702;S1C1;S1C2  56C          0.2, 0.2
S2C1;S2C2;  B4MU1702    B4MU1702;S2C1;S2C2  56C           0.1,0
S3C1;S3C2;  B4MU1702    B4MU1702;S3C1;S3C2  56C          0.1,0.2
S4C1;S4C2;  B4MU1702    B4MU1702;S4C1;S4C2  56C          0.1,0.1

我尝试过这样的事情：

df1.loc[df1.apply(lambda row: row.conc in [df2.conc.values], axis=1),
   df1['Delta']] = df1['Delta']+df2['Delta']

它给我错误

ValueError ：（'具有多个元素的数组的真值不明确。请使用a.any（）或a.all（）'，'发生在索引0'）

Answer 1

您可以通过set_index创建映射系列，然后通过pd.Series.apply使用自定义函数。这效率不高，但也不能保存用逗号分隔的代表数字数据的字符串。

请注意，f字符串需要Python 3.6+，如果需要，您可以改用str.format。

d = df1.set_index('conc')['Delta'].to_dict()

def get_vals(x):
    pre, *post = x.split(';')
    return ', '.join([str(d[f'{pre};{suffix}']) for suffix in post])

df2['Delta'] = df2['conc'].apply(get_vals)

print(df2[['conc', 'Delta']])

                 conc     Delta
0       B4MU1241;S1C1       0.2
1       B4MU1241;S2C1       0.2
2       B4MU1241;S3C1       1.0
3       B4MU1241;S4C1      11.1
4  B4MU1702;S1C1;S1C2  0.2, 0.2
5  B4MU1702;S2C1;S2C2  0.1, 0.0
6  B4MU1702;S3C1;S3C2  0.1, 0.2
7  B4MU1702;S4C1;S4C2  0.1, 0.1

Answer 2

这是另一种方法：

mapping = df1[['conc', 'Delta']].set_index('conc')['Delta'].to_dict()

df2['Delta'] = df2['conc'].apply(lambda x: [mapping[';'.join((x.split(';')[0], i))] for i in x.split(';')[1:]])

df2
#         Cell  NodeName Temparature-DUW                conc       Delta
#0       S1C1;  B4MU1241             60C       B4MU1241;S1C1       [0.2]
#1       S2C1;  B4MU1241             60C       B4MU1241;S2C1       [0.2]
#2       S3C1;  B4MU1241             60C       B4MU1241;S3C1       [1.0]
#3       S4C1;  B4MU1241             60C       B4MU1241;S4C1      [11.1]
#4  S1C1;S1C2;  B4MU1702             56C  B4MU1702;S1C1;S1C2  [0.2, 0.2]
#5  S2C1;S2C2;  B4MU1702             56C  B4MU1702;S2C1;S2C2  [0.1, 0.0]
#6  S3C1;S3C2;  B4MU1702             56C  B4MU1702;S3C1;S3C2  [0.1, 0.2]
#7  S4C1;S4C2;  B4MU1702             56C  B4MU1702;S4C1;S4C2  [0.1, 0.1]

基于条件的熊猫vlookup

2 个答案: