How to create a function with Plotly

Question

I am unable to apply a custom function to a data frame without receiving the error message "KeyError: 'state'". See code below.

import plotly.express as px

df = pd.DataFrame({'state':['IL', 'MN', "MN"]})

state_ct = (df
            .groupby(['state'])['state']
            .size()
            .reset_index(name='count'))


def heat_map(df):

    fig = px.choropleth(df,
                        locations=df['state'], 
                        locationmode='USA-states',
                        scope='usa',
                        color='count',
                        color_continuous_scale='reds')

    return(fig.show())

state_ct.apply(heat_map)

However, I can run the plotly code outside of the function with the state_ct data frame and have no issues. What is causing the issue when trying to run this through the custom Python function?

fig = px.choropleth(state_ct,
                    locations=state_ct['state'], 
                    locationmode='USA-states',
                    scope='usa',
                    color='count',
                    color_continuous_scale='reds')

fig.show()

Rob Raymond · Accepted Answer · 2021-08-12 21:23:31Z

1

https://pandas.pydata.org/docs/reference/api/pandas.DataFrame.apply.html
pandas apply() returns a dataframe or series. Your function returns a plotly figure.
your call context is standard python not a function to apply to each series in a dataframe
hence you should code as below

import plotly.express as px

df = pd.DataFrame({'state':['IL', 'MN', "MN"]})

state_ct = (df
            .groupby(['state'])['state']
            .size()
            .reset_index(name='count'))


def heat_map(df):

    fig = px.choropleth(df,
                        locations=df['state'], 
                        locationmode='USA-states',
                        scope='usa',
                        color='count',
                        color_continuous_scale='reds')

    return(fig.show())

heat_map(state_ct)

answered Aug 12, 2021 at 21:23

Rob Raymond

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