从每行的框架中提取唯一值，并将其添加到新列中

Question

我想解决以下问题：

假设我有一个datafram，看起来像这样

match              0              1              2        3    4    5    6    7
1           Morocco         France           Morocco      NaN  NaN  NaN  NaN  NaN
2           Morocco         France           Morocco      NaN  NaN  NaN  NaN  NaN
3           Morocco         France           NaN          NaN  NaN  NaN  NaN  NaN
4           China           United States    NaN          NaN  NaN  NaN  NaN  NaN
5           China           NaN              NaN          NaN  NaN  NaN  NaN  NaN

我正在寻找一种方法来查找每行中的唯一值，并将它们添加到另一列中，同时删除所有NaN

输出应如下所示：

match       8
1           Morocco, France
2           Morocco, France
3           Morocco, France
4           China, United States
5           China

对如何解决这个问题有什么建议吗

Answer 1

下面是在lambda中组合set和list的尝试：

df_ex[8] = [x for x in df_ex[[0,1,2,3,4,5,6,7]].values.tolist()]
df_ex[8] = df_ex[8].apply(lambda x: list(set([y for y in x if str(y)!='nan'])))

输出：

0         [Morocco, France]
1         [Morocco, France]
2         [Morocco, France]
3    [United States, China]
4                   [China]

Answer 2

# Convert each column dtype to str: x.astype(str)
# Null dtype became 'nan' so remove it: replace('nan', "")
# Concatenate each row entry: sum()
# Convert it to set to delete duplicate entries 
# Convert it to list to concatenate with "," as a string

df_new = df.apply(lambda x: ",".join(list(set(((x.astype(str)).sum()).replace('nan', "")))), axis=1)

Answer 3

使用：

cols = df.columns[df.columns.str.isnumeric()]
#or selecting columns
#cols = df.columns[1:]
#cols = df.columns.difference(['match'])
df[int(cols[-1])+1]=df[cols].agg(lambda x: ', '.join(set(x.dropna())),axis=1)
#for string type
#df[f'{int(cols[-1])+1}']=df[cols].stack().groupby(level=0).agg(', '.join)
df = df.reindex(columns = df.columns.difference(cols))
print(df)

                      8  match
0       France, Morocco      1
1       France, Morocco      2
2       France, Morocco      3
3  China, United_States      4
4                 China      5

我们还可以使用：

df[int(cols[-1])+1] = (df[cols].stack()
                               .groupby(level=0)
                               .agg(lambda x: ', '.join(set(x)),axis=1))