英文:
How can I rearrange df as the nodes index in pytorch geometric manner?
问题
我想按照PyTorch Geometric的方式重新排列我的DataFrame（data），将节点索引转换为原始名称以提取节点嵌入。
以下是代码的翻译部分：
import pandas as pddata = {'Source': ['Rainfall', 'SP2', 'SP2', 'Inflow', 'Rainfall', 'Inflow', 'Inflow', 'Inflow', 'SWT', 'SP1', 'SP1', 'SWD'],'Target': ['SP1', 'Evp', 'Outflow', 'SP2', 'SWD', 'SWD', 'SP2', 'SP1', 'SP1', 'SP2', 'Evp', 'Loss']}df = pd.DataFrame(data)nodes = pd.concat([df['Source'], df['Target']]).unique()node_indices = {node: i for i, node in enumerate(nodes)}df['Source'] = df['Source'].map(node_indices)df['Target'] = df['Target'].map(node_indices)
这是我的预期输出：
预期输出：
感谢任何意见或建议。
英文:
I'd like to rearrange my dataframe (data) as the node index to the original name in pytorch geometric manner for extracting node embedding.
import pandas as pddata = {'Source': ['Rainfall', 'SP2', 'SP2', 'Inflow','Rainfall','Inflow', 'Inflow', 'Inflow','SWT','SP1','SP1','SWD'],'Target': ['SP1', 'Evp', 'Outflow', 'SP2','SWD','SWD', 'SP2','SP1','SP1','SP2','Evp','Loss']}df = pd.DataFrame(data)nodes = pd.concat([df['Source'], df['Target']]).unique()node_indices = {node: i for i, node in enumerate(nodes)}df['Source'] = df['Source'].map(node_indices)df['Target'] = df['Target'].map(node_indices)
This is my expected outputs
Expected outputs:
Appreciate any though or suggestions.
答案1
得分: 0
我假设目标是获得一些参数的中间编码，并制作源值和目标值之间关系的可视化表示（似乎都来自同一个池，即此案例中的“节点”）。
data = {'Source': ['Rainfall', 'SP2', 'SP2', 'Inflow', 'Rainfall', 'Inflow', 'Inflow', 'Inflow', 'SWT', 'SP1', 'SP1', 'SWD'],'Target': ['SP1', 'Evp', 'Outflow', 'SP2', 'SWD', 'SWD', 'SP2', 'SP1', 'SP1', 'SP2', 'Evp', 'Loss']}df = pd.DataFrame(data)# 获取数据的所有唯一值nodes = {name: code for code, name in enumerate({*df.values.flat})}# 用它们的编码替换值df_coded = df.replace(nodes)# 将原始数据和编码数据连接到一个DataFrame中# 使用多级标头以便按名称和代码分别分隔列df_repr = pd.concat([df, df_coded], axis=1, keys=['Name','Code'])# 通过名称和代码重排列列df_repr = df_repr.iloc[:, [0, 2, 3, 1]]# 显示转置表示，隐藏原始索引print(df_repr.T.to_string(header=False))
通过这个代码，我们获得了以下输出：
更新
在行df_repr = df_repr.iloc[:, [0,2,3,1]]之后，这是按其索引重新排列列的方式。这里[0,2,3,1]表示“将第二列（索引为1）放在最后”。我们也可以使用.loc来实现，只需按所需顺序传递列的名称。在这种情况下，使用.loc会更长一些：
df_repr = df_repr.loc[:, [('Name', 'Source'),('Code', 'Source'),('Code', 'Target'),('Name', 'Target')]]
但如果可读性是首要任务，那当然最好使用loc而不是iloc。
P.S.因为我们这里只操作列，所以下面的方法也有效：
reordered_columns = [('Name', 'Source'),('Code', 'Source'),('Code', 'Target'),('Name', 'Target')]df_repr = df_repr[reordered_columns]
英文:
I'm assuming the goal is to get an intermediate encoding of some parameters and make a visual representation of the relationships between source and target values (which seem to be all from the same pool, i.e. nodes in this case).
data = {'Source': ['Rainfall', 'SP2', 'SP2', 'Inflow','Rainfall','Inflow', 'Inflow', 'Inflow','SWT','SP1','SP1','SWD'],'Target': ['SP1', 'Evp', 'Outflow', 'SP2','SWD','SWD', 'SP2','SP1','SP1','SP2','Evp','Loss']}df = pd.DataFrame(data)# get all unique values of datanodes = {name: code for code, name in enumerate({*df.values.flat})}# replace values with their codesdf_coded = df.replace(nodes)# connect original and encoded data in one DataFrame# use multilevel headers to ease separate columns by names and codesdf_repr = pd.concat([df, df_coded], axis=1, keys=['Name','Code'])# rearange columns like (source|name, source|code, target|code, target|name)df_repr = df_repr.iloc[:, [0,2,3,1]]# display transposed representation with hidden original indexesprint(df_repr.T.to_string(header=False))
With this we obtain the following output:
Update
As of the line df_repr = df_repr.iloc[:, [0,2,3,1]], this is reordering of columns by their indexes. Here [0,2,3,1] means put the second column (which is indexed by 1) at the very end. We can do it with .loc as well by passing names of columns in a desired order. It's just gonna be somewhat longer in this case:
df_repr = df_repr.loc[:, [('Name','Source'),('Code','Source'),('Code','Target'),('Name','Target')]]
But if readability is a priority, then of course it's better to use loc instead of iloc.
P.S. Because we are manipulating here only with columns, the following also works:
reordered_columns = [('Name','Source'),('Code','Source'),('Code','Target'),('Name','Target')]df_repr = df_repr[reordered_columns]
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