Pandas: Cannot plot a column against itself

[danielwlogan]

Code Sample, a copy-pastable example if possible

import pandas as pd
df = pd.DataFrame({'x': [1, 2], 'y': [2, 1]})
df.plot(x='x', y='x')

# A similar issue that should be tested once fixed
s1 = pd.Series(range(5), name="x")
s2 = pd.Series(range(10, 15), name="x")  # The following would work with "y" instead
pd.concat([s1, s2], axis=1).plot.scatter(x=0, y=1)

Problem description

The code above produces a KeyError: 'x'

Expected Output

I would expect the above code to produce a line plot that starts at (1, 1) and ends at (2, 2).

Output of pd.show_versions()

[paste the output of pd.show_versions() here below this line]

INSTALLED VERSIONS

commit: None
python: 3.6.5.final.0
python-bits: 64
OS: Linux
OS-release: 4.15.0-29-generic
machine: x86_64
processor: x86_64
byteorder: little
LC_ALL: None
LANG: en_CA.UTF-8
LOCALE: en_CA.UTF-8

pandas: 0.23.1
pytest: None
pip: 10.0.1
setuptools: 39.2.0
Cython: 0.28.3
numpy: 1.14.3
scipy: 1.1.0
pyarrow: None
xarray: None
IPython: 6.4.0
sphinx: None
patsy: 0.5.0
dateutil: 2.7.3
pytz: 2018.3
blosc: None
bottleneck: None
tables: None
numexpr: None
feather: None
matplotlib: 2.2.2
openpyxl: None
xlrd: None
xlwt: None
xlsxwriter: None
lxml: None
bs4: None
html5lib: 1.0.1
sqlalchemy: 1.2.7
pymysql: None
psycopg2: 2.7.4 (dt dec pq3 ext lo64)
jinja2: 2.10
s3fs: None
fastparquet: None
pandas_gbq: 0.4.1
pandas_datareader: None

[otro-mas1]

Two changes:
first change,
df.plot(x='x', y='x') ==> df.plot(x='x', y='y')

second change, if you expect a line from (1,1) to (2,2) ==> change series
df = pd.DataFrame({'x': [1, 2], 'y': [1, 2]})
df.plot(x='x', y='y')

[danielwlogan]

Thank you for the reply. I have some questions for you.

1. In your "first change", I don't want to plot y vs x, I want to plot x vs x. So I don't see how that accomplishes what I am looking for.

2. In your "second change", I do not want to create a new dataframe, I already have valid dataframe and I am simply selecting the columns that I want to plot. There doesn't appear anything in the documentation that would prevent passing the same column name to both the x and the y argument.

I understand that there is a work around (create a copy of the series I want to plot against itself) but it seems that the plot method should be able to handle the plotting without this step.

[otro-mas1]

ok, in that situation I would use the copy.
Good luck

[RobMulla]

I'm working on a PR that should solve this issue. The plotting method sets 'x' to the index by default. These two lines will allow for x == y

if y == x:
    data[y] = data.index

I've confirmed it works locally

[danielwlogan]

Yes, that output is exactly what I would expect. I figured it was something "simple" but hidden in the inner workings of plot(). Hopefully, your PR will make it into the next release. Thanks for tracking down a/the solution.

[WillAyd]

Just out of curiosity what is the use case for this? This is always just going to be a straight line no? Any practical application?

[RobMulla]

Not sure what @danielwlogan is intending to plot. It will always be a straight line, but if you used style='*' you could see where points sit on the line.

Now that I'm thinking about it, the fix won't work for a situation where df.plot(x='x', y = ['x', 'y']) which it probably should.

Either way, I don't think raising a KeyError: 'x' is the best output.

[danielwlogan]

There have been a number of times where this is useful. Most recently this is the situation that I am trying visualize.

I am trying to compare two models, f(t) and g(t). Say f(t) is what is currently done and I want to investigate if g(t) is a better model (e.g. gives a higher value). One way to visualize this is to scatter plot g(t) vs f(t) where each point is a different t. By plotting f(t) vs f(t) it is really clear if the point (f(t1), g(t1)) is above or below that line.

A related example is if you have a known value, x and an estimate of that value, x', then plotting the points (x, x') and overplotting the line x = x again visually shows how good the estimates are.

Furthermore, I believe the current plot() behavior is unnecessarily restrictive.

Good point @RobMulla re: the y list.

[ishanb97]

is this closed now?

[TomAugspurger]

No. There's a stalled PR https://github.com/pandas-dev/pandas/pull/22105

[Remnan13]

Directed here by Triage, but should I focus on jumping to #22105 instead? The self-join bug seems to have been resolved (at least addressed directly) in other issue threads!