How to style upper left cell of a Pandas DataFrame

Question

I understand to access the top left cell of a dataframe we need to use df.columns.name and I can see the pandas document on styling provides example to style row/column headers with apply_index (https://pandas.pydata.org/docs/user_guide/style.html)

The question I have is how to style this top left cell, say color it blue. Thanks.

import pandas as pd
d = {'col1': [1, 2], 'col2': [3, 4]}
df = pd.DataFrame(data=d)
df.columns.name = 'Test'
df

Update: ouroboros1’s answer below is very informative and helpful. But seems like when converting the Styler object to Excel, i.e. to_excel(), the format to the top left cell is not preserved.

In the documentation (and in Export to Excel section), it is stated that "Table level styles, and data cell CSS-classes are not included in the export to Excel: individual cells must have their properties mapped by the Styler.apply and/or Styler.applymap methods".

Update: Thanks to @ouroboros1
Here is a possible solution for anyone who is interested.

d = {'col1': [1, 2], 'col2': [3, 4]}
color_matrix_df = pd.DataFrame([['background-color:yellow', 'background-color:yellow'],
                ['background-color:yellow', 'background-color:blue']])
df = pd.DataFrame(data=d)
df.columns.name = 'Test'
df

def colors(df, color_matrix_df):
    style_df = pd.DataFrame(color_matrix_df.values, index=df.index, columns=df.columns)
    return style_df.applymap(lambda elem: elem)

df_upper_left_cell = df.style.set_table_styles(
    [{'selector': '.index_name',
      'props': [('background-color', 'IndianRed'),
                ('color', 'white')]
     }]
)

df_upper_left_cell.apply(colors, axis=None, color_matrix_df=color_matrix_df)

w = pd.ExcelWriter('Test.xlsx', engine='xlsxwriter')
df_upper_left_cell.to_excel(w, index=True)
wb = w.book
ws = w.sheets['Sheet1']
fmt_header = wb.add_format({'fg_color': '#cd5c5c', 'align': 'center'})
ws.write(0,0, df_upper_left_cell.data.columns.name, fmt_header)
w.save()

The above code will color the dataframe as shown below and saved the same to the Excel file.

Asked By: Chenyang

||

Source

Answer 1

Based on The Coding Bot: How to colour a specific cell in pandas dataframe based on its position?

You can use style.apply together with an iloc for that specific cell to assign a color.

def styling_specific_cell(x,row_idx,col_idx):
    #color = 'background-color: lightblue; color: blue'
    #color = 'color: blue'
    color = 'background-color: lightblue'
    df_styler = pd.DataFrame('', index=x.index, columns=x.columns)
    df_styler.iloc[row_idx, col_idx] = color
    return df_styler

style_idx_row = 0  
style_idx_column = 0   

df.style.apply(styling_specific_cell, row_idx = style_idx_row, col_idx = style_idx_column,
               axis = None)

Change the # within the function for font color or both.

To see the steps within the function check:

print(pd.DataFrame('', index=df.index, columns=df.columns))
print(df.iloc[0,0])

To assing a color to a specific index you can use style.applymap_index:

def color_b(v):
    return "background-color: lightblue;" if v == 0 else None

df.style.applymap_index(color_b)

Answered By: MagnusO_O

Answer 2

You can use Styler.set_table_styles for this.

We have the following options:

Top left corner (hereafter: "cell") is empty (no name set)
cell contains either df.columns.name or df.index.name, or both

Grouped there are three options to style cell (entirely, or partially). I’ll walk through them below. First the data:

import pandas as pd
df = pd.DataFrame({'col1': [1, 2], 'col2': [3, 4]})
df

Standard look:

Option 1. Cell is empty

Trick is to look at the output for print(df.style.set_table_styles().to_html()):

  <thead>
    <tr>
      <th class="blank level0" >&nbsp;</th>
      <th id="T_d7719_level0_col0" class="col_heading level0 col0" >col1</th>
      ...
    </tr>
  </thead>

Notice: <th class="blank level0" > </th>. We can access these class names and set the format. E.g. in this case, we access blank (level0 applying to all of level0, i.e. all cols and index too).

df.style.set_table_styles(
    [{'selector': '.blank',
      'props': [('background-color', 'IndianRed'),
                ('color', 'white')]
     }]
)

Result:

Option 2. Either df.columns.name or df.index.name, or both. One style to entire cell.

In either scenario, we will need .index_name instead of .blank. E.g.

df.columns.name = 'columns'

df.style.set_table_styles(
    [{'selector': '.index_name',
      'props': [('background-color', 'IndianRed'),
                ('color', 'white')]
     }]
)

Result:

Option 3. df.columns.name and df.index.name, different styles.

This one is a little more tricky, as we will need to add tr:nth-child(1) for columns.name and tr:nth-child(2) for index.name.

# just setting col and index name immediately inside `pd.DataFrame`
df = pd.DataFrame([[1,3],[2,4]], 
                  columns=pd.Index(['col1','col2'], name='columns'), 
                  index=pd.Index([0,1], name='index'))

df.style.set_table_styles(
    [{'selector': 'tr:nth-child(1) .index_name',
      'props': [('background-color', 'IndianRed'),
                ('color', 'white')]
     }, {'selector': 'tr:nth-child(2) .index_name',
      'props': [('background-color', '#EECACA'),
                ('color', 'black')]
     }]
)

Answered By: ouroboros1

How to style upper left cell of a Pandas DataFrame

Question:

Answers: