How to create subplots from each column in a pandas dataframe

Question

I have a dataframe 'df' with 36 columns, these columns are plotted onto a single plotly chart and displayed in html format using the code below.

import plotly.offline as py
import plotly.io as pio

pio.write_html(py.offline.plot([{
'x': df.index,
'y': df[col],
'name': col
}for col in trend_data.columns], filename=new_file_path))

I want to iterate through each column and create a subplot for each one. I have tried;

from plotly.subplots import make_subplots

sub_titles = df.columns()
fig = make_subplots(rows=6, cols=6, start_cell="bottom-left", subplot_titles=sub_titles)
for i in df.columns:
    fig.add_trace(i)

I created 6 rows and columns as that would give 36 plots and tried to use the header names as subplot titles but I get a ValueError stating it was expecting a 2d list of dictionaries.

Also, I have tried to add subplot titles by;

sub_titles = list(df)
fig = py.subplots.make_subplots(rows=6, cols=6, sub_titles=sub_titles)

This also returns an error. Any help is appreciatted.

vestland · Accepted Answer · 2019-11-12 10:04:28Z

7

Plot:

Code:

# imports
from plotly.subplots import make_subplots
import plotly.graph_objs as go
import pandas as pd
import numpy as np

# data
np.random.seed(123)
frame_rows = 50
n_plots = 36
frame_columns = ['V_'+str(e) for e in list(range(n_plots+1))]
df = pd.DataFrame(np.random.uniform(-10,10,size=(frame_rows, len(frame_columns))),
                  index=pd.date_range('1/1/2020', periods=frame_rows),
                    columns=frame_columns)
df=df.cumsum()+100
df.iloc[0]=100

# plotly setup
plot_rows=6
plot_cols=6
fig = make_subplots(rows=plot_rows, cols=plot_cols)

# add traces
x = 0
for i in range(1, plot_rows + 1):
    for j in range(1, plot_cols + 1):
        #print(str(i)+ ', ' + str(j))
        fig.add_trace(go.Scatter(x=df.index, y=df[df.columns[x]].values,
                                 name = df.columns[x],
                                 mode = 'lines'),
                     row=i,
                     col=j)

        x=x+1

# Format and show fig
fig.update_layout(height=1200, width=1200)
fig.show()

Addition: 1-column solution:

Code:

# imports
from plotly.subplots import make_subplots
import plotly.graph_objs as go
import pandas as pd
import numpy as np

# data
np.random.seed(123)
frame_rows = 50
frame_columns = ['V_'+str(e) for e in list(range(1,37))]
df = pd.DataFrame(np.random.uniform(-8,10,size=(frame_rows, len(frame_columns))),
                  index=pd.date_range('1/1/2020', periods=frame_rows),
                    columns=frame_columns)
df=df.cumsum()+100
df.iloc[0]=100

# plotly setup
plot_rows=6
plot_cols=6

lst1 = list(range(1,plot_rows+1))
lst2 = list(range(1,plot_cols+1))

fig = make_subplots(rows=36, cols=1, subplot_titles=df.columns, insets=[{'l': 0.1, 'b': 0.1, 'h':1}])

# add traces
x = 1
for i in lst1:
    for j in lst2:
        #print(str(i)+ ', ' + str(j))
        fig.add_trace(go.Scatter(x=df.index, y=df[df.columns[x-1]].values,
                                 name = df.columns[x-1],
                                 mode = 'lines',
                                 ),

                      row=x,
                     col=1)

        x=x+1

fig.update_layout(height=12000, width=1200)

fig.show()

Plots:

edited Nov 12, 2019 at 10:04

answered Oct 30, 2019 at 9:50

vestland

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5 Comments

Iceberg_Slim Over a year ago

Thanks for this, is it possible to make the graphs larger then scroll through the page to view them in more detail? I can' think of a way to fo this

vestland Over a year ago

@Jay That depends on where you are producing the plots. A Jupyter Notebook perhaps?

Iceberg_Slim Over a year ago

I'm using PyCharm to produce the plots. Did you just adjust fig.update_layout to have a huge length?

vestland Over a year ago

@Jay Yes, I did : fig.update_layout(height=12000, width=1200)

Bill Over a year ago

I'm wondering whether to learn plotly after using matplotlib and this surprised me. There is nothing as simple as fig, axes = plt.subplots(36, 1); for ax, col in zip(axes, df): df[col].plot(ax=ax) in the plotly API?

Albo · Accepted Answer · 2019-10-30 09:44:21Z

4

See documentation on how to use subplots. This could work:

UPDATE including subplot titles

fig = py.subplots.make_subplots(rows=36, cols=1, subplot_titles=df.columns)
j = 1
for i in df.columns:
    fig.add_trace(
        go.Scatter(
            {'x': df.index, 
             'y': df[i]}), 
             row=j, col=1)
    j += 1

This results in the following plot (with my data):

df = pd.DataFrame(np.random.randint(5, size=(5, 3)), columns=['one', 'two', 'three'])

edited Oct 30, 2019 at 9:44

answered Oct 30, 2019 at 8:48

Albo

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3 Comments

Iceberg_Slim Over a year ago

Thats works well! If i wanted to assign the column headers are subplot names, how would I do so? I tried; sub_titles = list(df) fig = py.subplots.make_subplots(rows=6, cols=6, sub_titles=sub_titles)

vestland Over a year ago

@Jay Did you want this in a 6x6 setup?

Iceberg_Slim Over a year ago

Thanks. @vestland Ideally I would like to be able to have 1 column but when I do this the graphs are tiny and not viewable. Is it possible to make the resulting html page scrollable?

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