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efueyo
Asked: 2020-05-19 09:43:06 +0800 CST 2020-05-19 09:43:06 +0800 CST

Decompose a dataframe into several dataframes, one for each column

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I have a dataframe df with the following structure

            Acatis  Avantage  TrueValue  ValorRelativo
Date                                                  
2014-07-31  199.08      10.0      10.22         11.631
2014-08-04  198.79      10.0      10.19         11.616
2014-08-05  198.59      10.0      10.17         11.614

I need to make, with each column, a dataframe whose name is the name of the column and whose name of the only column is "Close". Each of these dataframes must have this structure.

             Close
Date              
2014-07-31  199.08
2014-08-04  198.79
2014-08-05  198.59

I can do the following:

acatis = df[["Acatis"]].rename (columns = {"Acatis":"Close"})

but it solves the problem only partially, because the name of the new dataframe, acatis in this example, I have to write, that is, it is not associated by default in some way. On each execution of the program I can select another list which will result in other different names and I would have to modify the names of these new df's. I've tried pivot(), stack(), and unstack() and still can't find a solution. What can be the easiest way to do it? I will appreciate your help

python

1 Answers

  1. Best Answer

    Really what you are looking for is to create variables dynamically, this is possible in Python, although it is not generally recommended:

    • Using exec:

      for col in df.columns:
    exec(f'{col} = df[["{col}"]].rename(columns={{"{col}": "Close"}})')

      execit simply receives a string that is valid Python code and executes it. We just have to format the string appropriately.

    • Using the dictionaries directly globalsor localsdepending on the namespace where we want to create the variables:

      for col in df.columns:
    globals()[col] = df[[col]].rename(columns={col: "Close"})

      With this you have created a variable for each column, with the same name as the column, which refers to the new corresponding DataFrame:

      >>> Acatis
             Close
Date              
2014-07-31  199.08
2014-08-04  198.79
2014-08-05  198.59

    For various reasons (readability, complexity, intrinsic danger of modifying global variables dynamically, ...), the above is not recommended.

    • The correct and most commonly accepted way to do what you want is to use a dictionary, where the keys are the identifier of each DataFrame and the value is the DataFrame:

      dfs = {col: df[[col]].rename(columns={col: "Close"}) for col in df.columns}

      >>> dfs["Acatis"]

             Close
Date              
2014-07-31  199.08
2014-08-04  198.79
2014-08-05  198.59

    • If you want to access each DataFrame as an attribute, you should create your own namespace. You can use a class that acts as a container, with each DataFrame being attributes of it, or use types.SimpleNamespaces:

      from types import SimpleNamespace


dfs = SimpleNamespace(**{col: df[[col]].rename(columns={col: "Close"}) for col in df.columns})

      >>> dfs.Acatis

             Close
Date              
2014-07-31  199.08
2014-08-04  198.79
2014-08-05  198.59

    • You can do the same using a class as a container:

      class Dfs:
    pass

dfs = Dfs()

for col in df.columns:
    setattr(dfs, col, df[[col]].rename(columns={col: "Close"}))

    • Another solution, since you use Pandas, is to use a Pandas Series to store the DataFrames:

      dfs = pd.Series({col: df[[col]].rename(columns={col: "Close"}) for col in df.columns})

      >>> dfs.Acatis

             Close
Date              
2014-07-31  199.08
2014-08-04  198.79
2014-08-05  198.59

    Note

    If you want the identifier/key to be lowercase ( dfs["acatis"] / dfs.acatis) use str.lower:

    {col.lower(): df[[col]].rename ...}
    • 2