0

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

我想创建扫描df的列名并仅从列名中的列名中提取最后一个数字的逻辑，并将包含以下逻辑value的列的单元格中的值进行比较：value

  1. 如果列中的值value相等TRUE，则比较多值字典中的最后一个数字

  2. 使用多键字典中的第二个值来创建数据框列名

例子：

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
4

4 回答 4

1 
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]
于 2019-01-07T22:01:00.417 回答
0

哪里df，

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

利用：

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
于 2019-01-07T21:53:57.820 回答
0

国际大学联盟：

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
于 2019-01-07T22:06:17.753 回答
0 
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)
于 2019-01-07T22:02:05.907 回答