根据条件，如何用行不为空的列名填充列

Question

你好，我的问题和这篇文章几乎一样：How to fill in a column with column names whose rows are not NULL in Pandas?

但就我而言，我需要根据列名是国家还是细分来填充列，而不是进行串联。

编辑：表格 原来我有这个：

Segment	Country	Segment 1	Country 1	Segment 2
Nan	Nan	123456	123456	Nan
Nan	Nan	Nan	Nan	Nan
Nan	Nan	Nan	123456	123456
Nan	Nan	Nan	123456	123456

实际上我有这个（第一列由我代码中最后一行之前的两行填充：

Segment	Country	Segment 1	Country 1	Segment 2
Seg1 ; Country1 ;	Seg1 ; Country1 ;	123456	123456	Nan
Nan	Nan	Nan	Nan	Nan
country1 ; seg2 ;	country1 ; seg2 ;	Nan	123456	123456
country1 ; seg2 ;	country1 ; seg2 ;	Nan	123456	123456

我需要这个：

Segment	Country	Segment 1	Country 1	Segment 2
Segment 1	Country1	123456	123456	Nan
Nan	Nan	Nan	Nan	Nan
Segment 2	country1	Nan	123456	123456
Segment 2	country1	Nan	123456	123456

编辑：我的代码在尝试整合答案后实际上看起来像这样：错误是：AttributeError: Can only use .str accessor with string values!. Did you mean: 'std'?

#For each column in df, check if there is a value and if yes : first copy the value into the 'Amount' Column, then copy the column name into the 'Segment' or 'Country' columns
for column in df.columns[3:]:
    valueList = df[column][3:].values
    valueList = valueList[~pd.isna(valueList)]
    def detect(d):
        cols = d.columns.values
        dd = pd.DataFrame(columns=cols, index=d.index.unique())
        for col in cols:
            s = d[col].loc[d[col].str.contains(col[0:3], case=False)].str.replace(r'(\w+)(\d+)', col + r'\2')
            dd[col] = s
        return dd

    #Fill amount Column with other columns values if NaN
    if column in isSP:
        df['Amount'].fillna(df[column], inplace = True)
        df['Segment'] = df.iloc[:, 3:].notna().dot(df.columns[3:] + ';' ).str.strip(';')
        df['Country'] = df.iloc[:, 3:].notna().dot(df.columns[3:] + ' ; ' ).str.strip(';')
        df[['Segment', 'Country']] = detect(df[['Segment', 'Country']].apply(lambda x: x.astype(str).str.split(r'\s+[+]\s+').explode()))

非常感谢。

Answer 1

您可以使用以下解决方案。对于此解决方案，我首先定义了一个自定义函数，以根据与列名部分匹配的值过滤前两列，然后将它们替换为完整的列名：

def detect(d):
    cols = d.columns.values
    dd = pd.DataFrame(columns=cols, index=d.index.unique())
    for col in cols:
        s = d[col].loc[d[col].str.contains(col[0:3], case=False)].str.replace(r'(w+)(d+)', col + r'')
        dd[col] = s
    return dd

df[['Segment', 'Country']] = detect(df[['Segment', 'Country']].apply(lambda x: x.astype(str).str.split(';').explode()))
df

    Segment   Country Segment 1 Country 1 Segment 2
0  Segment1  Country1    123456    123456       Nan
1       NaN       NaN       Nan       Nan       Nan
2  Segment2  Country1       Nan    123456    123456
3  Segment2  Country1       Nan    123456    123456