Question

我喜欢合并Pandas Dataframe中长度不等的两列。

我尝试了很多方法进行合并，合并和联接，但是没有用。

keyList = ["Clone", "Chain", "Fragment", "R0", "R1", "R2"]
dataDict = {key: [] for key in keyList}
# Example for different list length
plist1 = ["ABCD", "DJFZ", "DHRZ"]
plist2 = ["ABCD", "DJFZ", "DHRZ", "JGJZ"]

filelist = ["E2_VH_Fab_R0.fasta", "E2_VH_scFV_R0.fasta", "E2_VH_Fab_R1.fasta", "E2_VH_scFV_R1.fasta","E2_VH_Fab_R2.fasta" ]

# Subsets are:
# E1 || E2 with VH || VL with Fab || scFV with R0 || R1 || R2 

for file in enumerate(filelist):
    # Get List with emits from class function
    peptidelist = clseq.processEmits()
    # Split filename into  6 parameters, see keylist
    fileparms = datafile.split('.')[0].split('_')

    # Iterate through peptide list and add the subsets into the dict
    for peptide in peptidelist:
        dataDict.setdefault("Clone", []).append(sclone)
        dataDict.setdefault("Chain", []).append(schain)
        dataDict.setdefault("Fragment", []).append(sfragment)
        # Set other Rounds as "NaN" to equal the length
        if "R0" in sround:
            dataDict.setdefault("R0", []).append(peptide)
            dataDict.setdefault("R1", []).append("NaN")
            dataDict.setdefault("R2", []).append("NaN")
        elif "R1" in sround:
            dataDict.setdefault("R0", []).append("NaN")
            dataDict.setdefault("R1", []).append(peptide)
            dataDict.setdefault("R2", []).append("NaN")
        elif "R2" in sround:
            dataDict.setdefault("R0", []).append("NaN")
            dataDict.setdefault("R1", []).append("NaN")
            dataDict.setdefault("R2", []).append(peptide)
        else:
            dataDict.setdefault("R0", []).append("NaN")
            dataDict.setdefault("R1", []).append("NaN")
            dataDict.setdefault("R2", []).append("NaN")

    dtframe.merge(pd.DataFrame(dataDict), on=['Clone', 'Chain',  'Fragment'], how='inner')

问题是，我的列表长度不同，我希望合并成一个数据框，其余的都用NaN填充。

此：

0    E2    VH      Fab  r0  nan
1    E2    VH      Fab  r0  nan
2    E2    VH      Fab  r0  nan
3    E2    VH      Fab  r0  nan
4    E2    VH      Fab  r0  nan
5    E2    VH      Fab  r0  nan

这：

0    E2    VH      Fab  nan  r1
1    E2    VH      Fab  nan  r1
2    E2    VH      Fab  nan  r1
3    E2    VH      Fab  nan  r1
4    E2    VH      Fab  nan  r1
5    E2    VH      Fab  nan  r1
6    E2    VH      Fab  nan  r1
7    E2    VH      Fab  nan  r1

应导致以下结果：

0     E2    VH      Fab  r0  r1
1     E2    VH      Fab  r0  r1
2     E2    VH      Fab  r0  r1
3     E2    VH      Fab  r0  r1
4     E2    VH      Fab  r0  r1
5     E2    VH      Fab  r0  r1
6     E2    VH      Fab  nan  r1
7     E2    VH      Fab  nan  r1

请注意，我所有的数据字段都是字符串。

Answer 1

这是combine_first。我们需要将索引设置为要合并的三列，然后为具有许多不同组的真实数据创建附加的cumcount级别。

df1['idx'] = df1.groupby(['Clone', 'Chain', 'Fragment']).cumcount()
df2['idx'] = df2.groupby(['Clone', 'Chain', 'Fragment']).cumcount()

df1 = df1.set_index(['Clone', 'Chain', 'Fragment', 'idx'])
df2 = df2.set_index(['Clone', 'Chain', 'Fragment', 'idx'])

df1.combine_first(df2).reset_index()
#  Clone Chain Fragment  idx   R0  R1
#0    E2    VH      Fab    0   r0  r1
#1    E2    VH      Fab    1   r0  r1
#2    E2    VH      Fab    2   r0  r1
#3    E2    VH      Fab    3   r0  r1
#4    E2    VH      Fab    4   r0  r1
#5    E2    VH      Fab    5   r0  r1
#6    E2    VH      Fab    6  NaN  r1
#7    E2    VH      Fab    7  NaN  r1

df1

  Clone Chain Fragment  R0  R1
0    E2    VH      Fab  r0 NaN
1    E2    VH      Fab  r0 NaN
2    E2    VH      Fab  r0 NaN
3    E2    VH      Fab  r0 NaN
4    E2    VH      Fab  r0 NaN
5    E2    VH      Fab  r0 NaN

df2

  Clone Chain Fragment  R0  R1
0    E2    VH      Fab NaN  r1
1    E2    VH      Fab NaN  r1
2    E2    VH      Fab NaN  r1
3    E2    VH      Fab NaN  r1
4    E2    VH      Fab NaN  r1
5    E2    VH      Fab NaN  r1
6    E2    VH      Fab NaN  r1
7    E2    VH      Fab NaN  r1

将数据框与现有值合并

1 个答案: