用不同的字典映射数据框的列

Question

我有一个包含两列的数据框架df

'data'列有一些值，例如:

df['data'] = [0, 1, 0, 1]

'Dict'列有多个字典，例如:

df['Dict'] = [{0: ['IPM_0'], 1: ['IPM_1']},
{0: ['SPM_0'], 1: ['SPM_1']},
{0: ['NPM_0'], 1: ['NPM_1']},
{0: ['DPM_0'], 1: ['DPM_1']}]

现在我想将每个'data'与'Dict'中的相对字典进行映射，例如结果看起来像:

df['result'] = [IPM_0, SPM_1, NPM_0, DPM_1]

我试图使用df['result'] = df.data.map(df.Dict)，但我得到一个列与df.Dict中相同的值。

我也尝试了df['result'] = df.data.apply(lambda rows: rows.map(rows.Dict))，但我得到一个错误'int' object has no attribute 'map'

我怎样才能达到这个结果?

起点:

data                          Dict
0     0  {0: ['IPM_0'], 1: ['IPM_1']}
1     1  {0: ['SPM_0'], 1: ['SPM_1']}
2     0  {0: ['NPM_0'], 1: ['NPM_1']}
3     1  {0: ['DPM_0'], 1: ['DPM_1']}

终点:

data                          Dict result
0     0  {0: ['IPM_0'], 1: ['IPM_1']}  IPM_0
1     1  {0: ['SPM_0'], 1: ['SPM_1']}  SPM_1
2     0  {0: ['NPM_0'], 1: ['NPM_1']}  NPM_0
3     1  {0: ['DPM_0'], 1: ['DPM_1']}  DPM_1

Answer 1

列表推导式总是很方便的:

df['result'] = [x[y][0] for x, y in zip(df['Dict'], df['data'])]

Answer 2

您可以使用apply。请记住，在数据框架中存储复杂的结构是没有效率的

!

df['result'] = df.apply(lambda r: r['Dict'][r['data']], axis=1)

输出:

data                          Dict   result
0     0  {0: ['IPM_0'], 1: ['IPM_1']}  [IPM_0]
1     1  {0: ['SPM_0'], 1: ['SPM_1']}  [SPM_1]
2     0  {0: ['NPM_0'], 1: ['NPM_1']}  [NPM_0]
3     1  {0: ['DPM_0'], 1: ['DPM_1']}  [DPM_1]

，如果你想要列表的第一项:

df['result'] = df.apply(lambda r: r['Dict'][r['data']][0], axis=1)

输出:

data                          Dict result
0     0  {0: ['IPM_0'], 1: ['IPM_1']}  IPM_0
1     1  {0: ['SPM_0'], 1: ['SPM_1']}  SPM_1
2     0  {0: ['NPM_0'], 1: ['NPM_1']}  NPM_0
3     1  {0: ['DPM_0'], 1: ['DPM_1']}  DPM_1