Search code examples
pythonpandasplotly

Plotly: bar plot with color red<0, green>0, divided by groups


Given a dataframe with 2 groups: (group1, group2), that have values > and < than 0: plot:

  • Bar plot
  • x = x
  • y = values, divided by group1, group2
  • color = red if value<0, green if value>0
  • legend shows group1, grou2 with different colors.

My current code however is not coloring as i would expect, and the legend is shown with the same color:

import pandas as pd
import numpy as np
import plotly.express as px 

df = pd.DataFrame( {
    "x" : [1,2,3],
    "group1" : [np.nan, 1, -0.5],
    "group2" : [np.nan, -0.2, 1],  
}).set_index("x")


df_ = df.reset_index().melt(id_vars = 'x')
fig = px.bar(df_, x='x', y='value', color='variable', barmode='group')
fig.update_traces(marker_color=['red' if val < 0 else 'green' for val in df_['value']], marker_line_color='black', marker_line_width=1.5)
fig.show()

OUT with indications of what i want to achieve: enter image description here


Solution

  • To stick with plotly.express, I would add a column to your dataframe, e.g. df_['positive'] with a boolean, and then color your plot by this variable.

    It would look like this:

    import pandas as pd
    import numpy as np
    import plotly.express as px 
    
    df = pd.DataFrame( {
        "x" : [1,2,3],
        "group1" : [np.nan, 1, -0.5],
        "group2" : [np.nan, -0.2, 1],  
    }).set_index("x")
    
    df_ = df.reset_index().melt(id_vars = 'x')
    df_['positive'] = (df_['value']>=0)
    
    fig = px.bar(df_, x='x', y='value',barmode = 'group',
                    color='positive',
                    color_discrete_map={
                        True: 'green',
                        False: 'red'
                        }
    )
    fig.update_traces(marker_line_color='black', marker_line_width=1.5)
    fig.show('browser')
    

    which yields the following : enter image description here

    EDIT following comments

    If you want to keep the colors AND the group distinction within plotly.express, one way could be to add patterns...

    Solution 1 : Every combination has its legend entry

    df = pd.DataFrame( {
        "x" : [1,2,3],
        "group1" : [np.nan, 1, -0.5],
        "group2" : [np.nan, -0.2, 1],  
    }).set_index("x")
    
    df_ = df.reset_index().melt(id_vars = 'x')
    positive = (df_['value']>=0)
    df_['positive'] = positive
    df_['sign'] = ['positive' if x else 'negative' for x in df_['positive']]
    # Each compbination of color and patterns
    fig = px.bar(df_, x='x', y='value',barmode = 'group',
                    color='sign',
                    color_discrete_map={
                        'positive': 'green',
                        'negative': 'red'
                        },
                    pattern_shape="variable")
    fig.update_layout(legend_title="Groups & Signs", bargap=0.5,bargroupgap=0.1)
    fig.show('browser')
    

    which outputs the following enter image description here

    Solution 2 : Legend only reflects patterns

    # Only patterns in legend
    fig = px.bar(df_, x='x', y='value', color='variable',
                    barmode='group',pattern_shape="variable")
    fig.update_layout(legend_title="Groups", bargap=0.5,bargroupgap=0.1)
    fig.for_each_trace(
        lambda trace: trace.update(marker_color=np.where(df_.loc[df_['variable'].eq(trace.name), 'value'] < 0, 'red', 'green'))
    )
    fig.show('browser')
    

    which outputs : enter image description here However I was not able to 'remove' the green color from the legend...