如何比较数据框列中的字符串值和单元格中的值以创建基于多值字典的新数据框？

Question

我有一个看起来像这样的 df:

id.1.value.1        id.2.value.2      id.1.question    id.2.value.2
TRUE                     FALSE             TRUE             TRUE

我想创建逻辑来扫描 df 的列名并仅从列名中具有 value 的列名中提取最后一个数字，然后比较包含 value 具有以下逻辑：

如果 value 列中的值等于 TRUE 则比较多值字典中的最后一个数字
使用多键字典中的第二个值来创建数据框列名

示例：

my_dict = {1: ('a', 'category'),2: ('b', 'category'),\
           3: ('c', 'category'),4:('d','category'),\
           5:('e','subcategory'),6:('f','subcategory'),\
           7:('g','subcategory'),8:('h','subcategory'),\
           9:('i','subcategory'),10:('j','subcategory'),\
           11:('k','subcategor'),12:('l','subcategory'),\
           13:('m','subcategory'),14:('n','subcategory'),\
           15:('o','subcategory'),16:('p','subcategory'),\
           17:('q','subcategory'),18:('r','subcategory'),\
           19:('s','subcategory'),20:('t','subcategory'),\
           21:('u','subcategory'),22:('v','subcategory'),\
           23:('w','subcategory'),24:('x','subcategory')

           }

如果我当前的 df 看起来像这样：

id.1.value.1        id.2.value.2      id.1.question    id.6.value.6
    TRUE                 FALSE             TRUE             TRUE

新的 df 应该是这样的：

category    subcategory
a               f

Answer 1

哪里 df,

   id.1.value.1  id.2.value.2  id.1.question  id.6.value.6
0          True         False           True          True

使用：

i = df.loc[:,df.columns[df.iloc[0]]].filter(like='value').columns.str.split('.').str[-1].astype(int).tolist()

my_dict = {1: ('a', 'category'),2: ('b', 'category'),\
           3: ('c', 'category'),4:('d','category'),\
           5:('e','subcategory'),6:('f','subcategory'),\
           7:('g','subcategory'),8:('h','subcategory'),\
           9:('i','subcategory'),10:('j','subcategory'),\
           11:('k','subcategor'),12:('l','subcategory'),\
           13:('m','subcategory'),14:('n','subcategory'),\
           15:('o','subcategory'),16:('p','subcategory'),\
           17:('q','subcategory'),18:('r','subcategory'),\
           19:('s','subcategory'),20:('t','subcategory'),\
           21:('u','subcategory'),22:('v','subcategory'),\
           23:('w','subcategory'),24:('x','subcategory')}

df1 = pd.DataFrame.from_dict(my_dict, orient='index')

df_out = df1.loc[i].set_index(1).T

print(df_out)

输出：

1 category subcategory
0        a           f

Answer 2

names = df.columns
new_df = pd.DataFrame()
for name in names:    
    if ('value' in name) & df[name][0]:
        last_number = int(name[-1])
        key, value = my_dict[last_number]
        try:
            new_df[value][0] = list(new_df[value][0]) + [key]
        except:
            new_df[value] = [key]

Answer 3

new_df = pd.DataFrame()

# get column names
for col in (list(df)):

    if "value" in col:

        try:
            # operate only in columns where a valid number is found
            value = df[col].rpartition('.')[:-1]

            # When df== True
            if df.loc[col,1]==True:
                new_df[my_dict[value][1]]= my_dict[value][0]
        except Exception as e:
            print(e)

Answer 4

IIUC:

ans = [my_dict[int(x[-1])] for x in df1.where(df1.loc[:,['value' in x for x in df1.columns]]).dropna(axis=1)]
pd.DataFrame.from_dict({v: k for k, v in dict(ans).items()}, orient='index').T

输出：

  category subcategory
0        a           f

如何比较数据框列中的字符串值和单元格中的值以创建基于多值字典的新数据框？

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