Question

我要解决以下问题：

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

我正在寻找一种在每一行中查找唯一值并将其添加到另一列，同时删除所有NaN的方法。

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

这是尝试在set中组合list和lambda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

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))

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})

每行从熊猫帧中提取唯一值并将其添加到新列中

4 个答案: