根据python中另一个选项卡文件中的条件从选项卡文件中删除行

Question

我有两个标签文件，例如： file1.txt

Clustername Seqname1 Seqname2
Cluster1 Seq1(+) SeqA
Cluster1 Seq2(-) SeqA
Cluster1 Seq3(+) SeqB
Cluster1 Seq300(+) SeqB
Cluster1 Seq90(+) SeqL
Cluster1 Seq90(+) SeqO
Cluster1 Seq2(-) SeqC
Cluster2 Seq8(-) SeqY
Cluster2 Seq8(-) SeqH
Cluster2 Seq8(-) SeqP
Cluster2 Seq79(-) SeqY
Cluster3 Seq10(+) SeqK
Cluster3 Seq10(+) SeqS
Cluster3 Seq10(+) SeqT
Cluster4 Seq300(+) SeqB

file2.txt

Clustername Names
Cluster1    SeqA
Cluster1    Seq1(+)
Cluster1    SeqC
Cluster1    Seq2(-)
Cluster1    SeqO
Cluster1    Seq3(+)
Cluster1    Seq90(+)
Cluster1    SeqB
Cluster1    SeqG
Cluster2    Seq8(-)
Cluster2    SeqY
Cluster2    SeqH
Cluster3    Seq10(+)
Cluster3    SeqK
Cluster4    SeqB
Cluster4    Seq300(+)

正如您在file2.txt 中看到的那样，Cluster1 中不存在 SeqL，那么我想删除该行： Cluster1 Seq90(+) SeqL 来自 file1.txt

Seq300(+) 在Cluster1 中也不存在，然后我删除该行：

Cluster1 Seq300(+) SeqB

来自 file1.txt

同样适用于：

Cluster2 Seq8(-) SeqP
Cluster2 Seq79(-) SeqY

file2.txt中的CLuster2中没有SeqP，Cluster2中也没有Seq79(-)，然后我删除行：

Cluster2 Seq8(-) SeqP
Cluster2 Seq79(-) SeqY

来自 file1.txt

同样适用于：

Cluster3 Seq10(+) SeqS
Cluster3 Seq10(+) SeqT

因为SeqS和SeqT不在file2.txt中的Cluster2中，所以我从file1.txt中删除以下两行：

 Cluster3 Seq10(+) SeqS
 Cluster3 Seq10(+) SeqT

最后我应该得到一个 ex file1.txt，例如：

Clustername Seqname1 Seqname2
Cluster1 Seq1(+) SeqA
Cluster1 Seq2(-) SeqA
Cluster1 Seq3(+) SeqB
Cluster1 Seq90(+) SeqO
Cluster1 Seq2(-) SeqC
Cluster2 Seq8(-) SeqY
Cluster2 Seq8(-) SeqH
Cluster3 Seq10(+) SeqK
Cluster4 Seq300(+) SeqB

Answer 1

使用DataFrame.merge + DataFrame.reindex 获取原始列：

new_df=( df1.merge(df2,left_on=['Clustername','Seqname1'],right_on=['Clustername','Names'])
            .merge(df2,left_on=['Clustername','Seqname2'],right_on=['Clustername','Names'])
            .reindex(columns=df1.columns))
print(new_df)

输出

  Clustername   Seqname1 Seqname2
0    Cluster1    Seq1(+)     SeqA
1    Cluster1    Seq2(-)     SeqA
2    Cluster1    Seq2(-)     SeqC
3    Cluster1    Seq3(+)     SeqB
4    Cluster1   Seq90(+)     SeqO
5    Cluster2    Seq8(-)     SeqY
6    Cluster2    Seq8(-)     SeqH
7    Cluster3   Seq10(+)     SeqK
8    Cluster4  Seq300(+)     SeqB

---

n seqnames 列的解决方案：

df1['aux']=df1.groupby('Clustername').cumcount()

new_df= ( df1.melt(['Clustername','aux'],var_name='Seq')
             .merge(df2,left_on=['Clustername','value'],right_on=['Clustername','Names'])
             .groupby(['Clustername','aux'])
             .filter(lambda x: x.value.size>=(len(df1.columns)-2))
             .pivot_table(index=['Clustername','aux'],columns='Seq',values='value',aggfunc=''.join)
             .reset_index()
             .drop('aux',axis=1)
             .rename_axis(columns=None) )

print(new_df)

输出

  Clustername   Seqname1 Seqname2
0    Cluster1    Seq1(+)     SeqA
1    Cluster1    Seq2(-)     SeqA
2    Cluster1    Seq3(+)     SeqB
3    Cluster1   Seq90(+)     SeqO
4    Cluster1    Seq2(-)     SeqC
5    Cluster2    Seq8(-)     SeqY
6    Cluster2    Seq8(-)     SeqH
7    Cluster3   Seq10(+)     SeqK
8    Cluster4  Seq300(+)     SeqB

Answer 2

创建一个包含所有必要值的列 df1 是file1.txt，df2 是file2.txt

df1['cs1'] = df1['Clustername'] + ' ' + df1['Seqname1']
df1['cs2'] = df1['Clustername'] + ' ' + df1['Seqname2']

df2['seq2'] = df2['Names'][~df2['Names'].str.contains('(\()')]

df2['cs1'] = df2['Clustername'] + ' ' + df2['Names']
df2['cs2'] = df2['Clustername'] + ' ' + df2['seq2']

result = df1[(df1['cs1'].isin(df2['cs1'])) & (df1['cs2'].isin(df2['cs2']))]

过滤所需的列 result[['Clustername', 'Seqname1', 'Seqname2']]

   Clustername  Seqname1 Seqname2
0     Cluster1   Seq1(+)     SeqA
1     Cluster1   Seq2(-)     SeqA
2     Cluster1   Seq3(+)     SeqB
5     Cluster1  Seq90(+)     SeqO
6     Cluster1   Seq2(-)     SeqC
7     Cluster2   Seq8(-)     SeqY
8     Cluster2   Seq8(-)     SeqH
11    Cluster3  Seq10(+)     SeqK
12    Cluster4  Seq300(+)    SeqB