从每行的熊猫帧中提取唯一值并将它们添加到新列中

Question

我有以下问题要解决：

假设我有一个看起来像这样的数据帧

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一行中的唯一值并将它们添加到另一列，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

有关如何解决此问题的任何建议？

Answer 1

# 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 2

这是在lambda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 3

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

我们也可以使用：

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

Answer 4

很长的路要走

dee = dict(tuple(df.groupby('Match')))
tmp = []
tmp2 = []
for k,v in dee.items():

    tmp.append(k)

    for i in v.columns.tolist():
        tmp3 = []
        #print(i)
        tmp3.append(str(v[i]))
    tmp2.append(tmp3)
new = pd.DataFrame({'Match':tmp,'List':tmp2})