pandas 融化功能 [英] Pandas Melt Function

问题描述

我有一个数据框:

df = pd.DataFrame([[2, 4, 7, 8, 1, 3, 2013], [9, 2, 4, 5, 5, 6, 2014]], columns=['Amy', 'Bob', 'Carl', 'Chris', 'Ben', 'Other', 'Year'])

   Amy  Bob  Carl  Chris  Ben  Other  Year
0    2    4     7      8    1      3  2013
1    9    2     4      5    5      6  2014

还有字典:

d = {'A': ['Amy'], 'B': ['Bob', 'Ben'], 'C': ['Carl', 'Chris']}

我想重塑我的数据框，使其看起来像这样:

I would like to reshape my dataframe to look like this:

    Group   Name  Year  Value
 0      A    Amy  2013      2
 1      A    Amy  2014      9
 2      B    Bob  2013      4
 3      B    Bob  2014      2
 4      B    Ben  2013      1
 5      B    Ben  2014      5
 6      C   Carl  2013      7
 7      C   Carl  2014      4
 8      C  Chris  2013      8
 9      C  Chris  2014      5
10  Other         2013      3
11  Other         2014      6

请注意，Other在Name列中没有任何值，并且行的顺序无关紧要.我想我应该使用melt函数，但是我遇到的示例不太清楚.

Note that Other doesn't have any values in the Name column and the order of the rows does not matter. I think I should be using the melt function but the examples that I've come across aren't too clear.

推荐答案

melt可以帮助您实现目标.

melt gets you part way there.

In [29]: m = pd.melt(df, id_vars=['Year'], var_name='Name')

除了Group之外，其他所有内容都包含在内.为此，我们还需要重新调整d的形状.

This has everything except Group. To get that, we need to reshape d a bit as well.

In [30]: d2 = {}

In [31]: for k, v in d.items():
    for item in v:
        d2[item] = k
   ....:

In [32]: d2
Out[32]: {'Amy': 'A', 'Ben': 'B', 'Bob': 'B', 'Carl': 'C', 'Chris': 'C'}

In [34]: m['Group'] = m['Name'].map(d2)

In [35]: m
Out[35]:
    Year   Name  value Group
0   2013    Amy      2     A
1   2014    Amy      9     A
2   2013    Bob      4     B
3   2014    Bob      2     B
4   2013   Carl      7     C
..   ...    ...    ...   ...
7   2014  Chris      5     C
8   2013    Ben      1     B
9   2014    Ben      5     B
10  2013  Other      3   NaN
11  2014  Other      6   NaN

[12 rows x 4 columns]

并将其他"从Name移至Group

In [8]: mask = m['Name'] == 'Other'

In [9]: m.loc[mask, 'Name'] = ''

In [10]: m.loc[mask, 'Group'] = 'Other'

In [11]: m
Out[11]:
    Year   Name  value  Group
0   2013    Amy      2      A
1   2014    Amy      9      A
2   2013    Bob      4      B
3   2014    Bob      2      B
4   2013   Carl      7      C
..   ...    ...    ...    ...
7   2014  Chris      5      C
8   2013    Ben      1      B
9   2014    Ben      5      B
10  2013             3  Other
11  2014             6  Other

[12 rows x 4 columns]

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