Python - Interactive multiple line graph with Plotly Express - need each line based on similar values in column

Question

Need to plot multiple lines (large dataset - will yield ~4,500 lines) on interactive 2-dimensional line graph using Plotly Express.

Problem is that my variables (x and y axes) are in 2 separate columns, and the number of data points for each line is different. The number of data points for each line will be based upon the 'API/UWI' column --- the rows where the 'API/UWI' values are constant will represent all the data points for 1 line.

i.e., when the value in the 'API/UWI' column changes, a new line starts.

An example of a small portion of my dataset is below for clarity.

Example dataset

In my first attempt, I separated the dataframe into multiple dataframes based on the unique values in the 'API/UWI' column and plotted all of those dataframes into graphs. It worked! However, it produced 4,500 graphs, rather than a single graph with 4,500 lines.

Is there a better way to accomplish this and produce a single graph with 4,500 lines?

I am posting my code below for the attempt I made above, along with an example of a graph that was produced (with a single line).

Please be detailed as possible on your solutions. This is my FIRST attempt at coding of any kind. I'm very much a beginner. PLEASE & THANKS!

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

    excel_file = r"C:\Users\kevin\Desktop\Bone_Spring_Data_2.xlsx"

    df = pd.read_excel(excel_file)

    split_values_API = df['API/UWI'].unique()

    for API in split_values_API:
        df1 = df[df['API/UWI'] == API]
        df1 = df1.sort_values(by="Monthly Production Date")
        GOR_data = px.line(df1, x='Monthly Production Date' , y='MONTHLY GOR')
        GOR_data.show()

Example plot with 1 line

*Edit in response to comment:

The output of df.head(21).to_dict() is as follows:

{'API/UWI': {0: 30015209400000,
  1: 30015209400000,
  2: 30015209400000,
  3: 30015209400000,
  4: 30015209400000,
  5: 30015209400000,
  6: 30015209400000,
  7: 30015209400000,
  8: 30015221570000,
  9: 30015221570000,
  10: 30015221570000,
  11: 30015221570000,
  12: 30015221620000,
  13: 30015221620000,
  14: 30015221620000,
  15: 30015221620000,
  16: 30015221620000,
  17: 30015221620000,
  18: 30015221620000,
  19: 30015221620000,
  20: 30015221620000},
 'Monthly Production Date': {0: Timestamp('2002-04-01 00:00:00'),
  1: Timestamp('2002-05-01 00:00:00'),
  2: Timestamp('2002-06-01 00:00:00'),
  3: Timestamp('2002-07-01 00:00:00'),
  4: Timestamp('2002-08-01 00:00:00'),
  5: Timestamp('2002-09-01 00:00:00'),
  6: Timestamp('2002-10-01 00:00:00'),
  7: Timestamp('2006-07-01 00:00:00'),
  8: Timestamp('2008-08-01 00:00:00'),
  9: Timestamp('2008-09-01 00:00:00'),
  10: Timestamp('2008-10-01 00:00:00'),
  11: Timestamp('2008-11-01 00:00:00'),
  12: Timestamp('2016-10-01 00:00:00'),
  13: Timestamp('2016-11-01 00:00:00'),
  14: Timestamp('2016-12-01 00:00:00'),
  15: Timestamp('2017-01-01 00:00:00'),
  16: Timestamp('2017-02-01 00:00:00'),
  17: Timestamp('2017-03-01 00:00:00'),
  18: Timestamp('2017-04-01 00:00:00'),
  19: Timestamp('2017-05-01 00:00:00'),
  20: Timestamp('2017-06-01 00:00:00')},
 'MONTHLY GOR': {0: 1.278688524590164,
  1: 0.8455284552845529,
  2: 1.8529411764705883,
  3: 0.736,
  4: 1.6818181818181819,
  5: 0.9795918367346939,
  6: 0.5303030303030303,
  7: 0.0,
  8: 14.523809523809524,
  9: 17.07622203811102,
  10: 16.334231805929917,
  11: 14.918367346938776,
  12: 1.4124008651766402,
  13: 1.8545081967213115,
  14: 1.2862351868732909,
  15: 1.4340557275541796,
  16: 2.2898674647285167,
  17: 2.7108673978065805,
  18: 14.311827956989248,
  19: 2.871877001921845,
  20: 2.8629370629370627}}

This is an example dataset to match my example (attached figure).

Hi @cowboykevin05, welcome on SO. Please read about mcve and how-to-ask. In particular it will be great if you can share the output of df.head(20).to_dict() instead a picture. — rpanai
– rpanai, Commented Jun 17, 2020 at 16:58
You should edit your own question as there is a limit for characters on comments. — rpanai
– rpanai, Commented Jun 17, 2020 at 17:13
@rpanai Thanks. I edited the post to answer your question. Note the comment at the bottom. I will read through the links you sent. — cowboykevin05
– cowboykevin05, Commented Jun 17, 2020 at 17:19
@rpanai I'm not sure why it's not showing in the post, but the last section of the output of df.head(20).to_dict() is supposed to say 'MONTHLY GOR': {0: 1.1925754060324827, — cowboykevin05
– cowboykevin05, Commented Jun 17, 2020 at 17:32

rpanai · Accepted Answer · 2020-06-17 17:49:13Z

1

First of all if you want to use plotly.express you might just use

px.line(df,
        x='Monthly Production Date',
        y='MONTHLY GOR',
        color='API/UWI')

the parameter color allows you to have a line for every different 'API/UWI' in your df.

If you want to use plotly.graph_objs instead you can do

import plotly.graph_objs as go
fig = go.Figure()
for ts in df.groupby("API/UWI"):
    ts = ts[1]
    fig.add_trace(go.Scatter(x=ts['Monthly Production Date'],
                             y=ts['MONTHLY GOR']))
fig.show()

In this case you are adding a trace for every "API/UWI".

answered Jun 17, 2020 at 17:49

rpanai

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3 Comments

cowboykevin05 Over a year ago

Thank you! This worked. The reason I chose to use Plotly Express is because it seemed to work really well for hovering over the data with your mouse and seeing additional specified information. I will still need to figure that out for the plotly.graph_objs as go method. If you know of a good source to find that information, please let me know. I still have difficulties putting my questions to words as a novice. Thanks again!

cowboykevin05 Over a year ago

also, I'm confused as to why there is an index value of 1 at ts = ts[1] . Can you explain that please?

rpanai Over a year ago

Whenever you take an element ts within a groupby this is a tuple where the first element ts[0]is the key for the groupby while the second one ts[1] is a dataframe. you could check printing out ts, ts[0] and ts[1] at the end of the loop.

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