'Scatter plot form dataframe with index on x-axis
I've got pandas DataFrame, df, with index named date and the columns columnA, columnB and columnC
I am trying to scatter plot index on a x-axis and columnA on a y-axis using the DataFrame syntax.
When I try:
df.plot(kind='scatter', x='date', y='columnA')
I ma getting an error KeyError: 'date' probably because the date is not column
df.plot(kind='scatter', y='columnA')
I am getting an error:
ValueError: scatter requires and x and y column
so no default index on x-axis.
df.plot(kind='scatter', x=df.index, y='columnA')
I am getting error
KeyError: "DatetimeIndex(['1818-01-01', '1818-01-02', '1818-01-03', '1818-01-04',\n
                          '1818-01-05', '1818-01-06', '1818-01-07', '1818-01-08',\n
                          '1818-01-09', '1818-01-10',\n               ...\n  
                          '2018-03-22', '2018-03-23', '2018-03-24', '2018-03-25',\n
                          '2018-03-26', '2018-03-27', '2018-03-28', '2018-03-29',\n 
                          '2018-03-30', '2018-03-31'],\n  
dtype='datetime64[ns]', name='date', length=73139, freq=None) not in index"
I can plot it if I use matplotlib.pyplot directly
plt.scatter(df.index, df['columnA'])
Is there a way to plot index as x-axis using the DataFrame kind syntax?
Solution 1:[1]
This is kind of ugly (I think the matplotlib solution you used in your question is better, FWIW), but you can always create a temporary DataFrame with the index as a column usinng
df.reset_index()
If the index was nameless, the default name will be 'index'. Assuming this is the case, you could use
df.reset_index().plot(kind='scatter', x='index', y='columnA')
Solution 2:[2]
A more simple solution would be:
df['x1'] = df.index
df.plot(kind='scatter', x='x1', y='columnA')
Just create the index variable outside of the plot statement.
Solution 3:[3]
At least in pandas>1.4 whats easiest is this:
df['columnA'].plot(style=".")
This lets you mix scatter and line plots, as well as use the standard pandas plot interface
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source | 
|---|---|
| Solution 1 | Ami Tavory | 
| Solution 2 | Clovis | 
| Solution 3 | Amal Duriseti | 
