Adding labels in x y scatter plot with seaborn

Question

I've spent hours on trying to do what I thought was a simple task, which is to add labels onto an XY plot while using seaborn.

Here's my code

import seaborn as sns
import matplotlib.pyplot as plt
%matplotlib inline

df_iris=sns.load_dataset("iris") 

sns.lmplot('sepal_length', # Horizontal axis
           'sepal_width', # Vertical axis
           data=df_iris, # Data source
           fit_reg=False, # Don't fix a regression line
           size = 8,
           aspect =2 ) # size and dimension

plt.title('Example Plot')
# Set x-axis label
plt.xlabel('Sepal Length')
# Set y-axis label
plt.ylabel('Sepal Width')

I would like to add to each dot on the plot the text in "species" column.

I've seen many examples using matplotlib but not using seaborn.

Any ideas? Thank you.

Can you provide an example data frame? Does `z` contain label information for both X and Y axes? Do you want to label the entire axis, or axis tick marks? Seaborn uses Matplotlib under the hood - are you saying that you do not want to use `plt` methods but `sns` methods only to label your plots? — andrew_reece, Sep 03 '17 at 20:53

score 57 · Accepted Answer · answered Sep 03 '17 at 23:30

One way you can do this is as follows:

import seaborn as sns
import matplotlib.pyplot as plt
import pandas as pd
%matplotlib inline

df_iris=sns.load_dataset("iris") 

ax = sns.lmplot('sepal_length', # Horizontal axis
           'sepal_width', # Vertical axis
           data=df_iris, # Data source
           fit_reg=False, # Don't fix a regression line
           size = 10,
           aspect =2 ) # size and dimension

plt.title('Example Plot')
# Set x-axis label
plt.xlabel('Sepal Length')
# Set y-axis label
plt.ylabel('Sepal Width')


def label_point(x, y, val, ax):
    a = pd.concat({'x': x, 'y': y, 'val': val}, axis=1)
    for i, point in a.iterrows():
        ax.text(point['x']+.02, point['y'], str(point['val']))

label_point(df_iris.sepal_length, df_iris.sepal_width, df_iris.species, plt.gca())

Thank you Scott. It does plot but for me the string that's plotted looks weird. Each point says something along the following: "species: setosa, Name: 3, dtype: object" Any idea how to fix that? — Trexion Kameha, Sep 04 '17 at 00:08

LucasBoatwright · Answer 2 · 2022-11-22T20:17:31.870

29

Here's a more up-to-date answer that doesn't suffer from the string issue described in the comments.

import seaborn as sns
import matplotlib.pyplot as plt
%matplotlib inline

df_iris=sns.load_dataset("iris") 

plt.figure(figsize=(20,10))
p1 = sns.scatterplot(x='sepal_length', # Horizontal axis
       y='sepal_width', # Vertical axis
       data=df_iris, # Data source
       size = 8,
       legend=False)  

for line in range(0,df_iris.shape[0]):
     p1.text(df_iris.sepal_length[line]+0.01, df_iris.sepal_width[line], 
     df_iris.species[line], horizontalalignment='left', 
     size='medium', color='black', weight='semibold')

plt.title('Example Plot')
# Set x-axis label
plt.xlabel('Sepal Length')
# Set y-axis label
plt.ylabel('Sepal Width')

edited Nov 22 '22 at 20:17

answered Feb 20 '19 at 14:55

LucasBoatwright

1,456
1
16
20

This logic assumes (by looping an iterator line through data[x][line]) that the dataframe has an incrementing index without any gaps. This will not be true, for example, with filtered dataframes. The function will raise a KeyError. – defraggled Apr 23 '21 at 12:25
User can workaround this problem by passing `df.reset_index(drop=True)` instead of the raw df. – defraggled Apr 23 '21 at 12:31
With my version of seaborn I had to add `x=` and `y=` to the arguments like so: ```sns.scatterplot(x='sepal_length', y='sepal_width', data=df_iris, size = 8, legend=False)``` – jss367 Nov 22 '22 at 19:44

Paul Rougieux · Answer 3 · 2020-10-02T14:51:37.000

Thanks to the 2 other answers, here is a function scatter_text that makes it possible to reuse these plots several times.

import seaborn as sns
import matplotlib.pyplot as plt

def scatter_text(x, y, text_column, data, title, xlabel, ylabel):
    """Scatter plot with country codes on the x y coordinates
       Based on this answer: https://stackoverflow.com/a/54789170/2641825"""
    # Create the scatter plot
    p1 = sns.scatterplot(x, y, data=data, size = 8, legend=False)
    # Add text besides each point
    for line in range(0,data.shape[0]):
         p1.text(data[x][line]+0.01, data[y][line], 
                 data[text_column][line], horizontalalignment='left', 
                 size='medium', color='black', weight='semibold')
    # Set title and axis labels
    plt.title(title)
    plt.xlabel(xlabel)
    plt.ylabel(ylabel)
    return p1

Use the function as follows:

df_iris=sns.load_dataset("iris") 
plt.figure(figsize=(20,10))
scatter_text('sepal_length', 'sepal_width', 'species',
             data = df_iris, 
             title = 'Iris sepals', 
             xlabel = 'Sepal Length (cm)',
             ylabel = 'Sepal Width (cm)')

See also this answer on how to have a function that returns a plot: https://stackoverflow.com/a/43926055/2641825

This logic assumes (by looping an iterator `line` through `data[x][line]`) that the dataframe has an incrementing index without any gaps. This will not be true, for example, with filtered dataframes. The function will raise a KeyError. — defraggled, Apr 23 '21 at 12:24
User can workaround this problem by passing `df.reset_index(drop=True)` instead of the raw df. — defraggled, Apr 23 '21 at 12:31

Maciej Skorski · Answer 4 · 2023-03-22T07:41:41.670

4

Use the powerful declarative API to avoid loops (seaborn>=0.12).

Specifically, put x,y, and annotations into a pandas data frame and call plotting.

Here is an example from my own research work.

import seaborn.objects as so
import pandas as pd
import matplotlib.pyplot as plt

df = pd.DataFrame(..,columns=['phase','P(X=1)','text'])

fig,ax = plt.subplots()
    p = so.Plot(df,x='phase',y='P(X=1)',text='text').add(so.Dot(marker='+')).add(so.Text(halign='left'))
    p.on(ax).show()

edited Mar 22 '23 at 07:41

answered Jan 09 '23 at 22:31

Maciej Skorski

2,303
6
14

1

Aside from omitting `import matplotlib.pyplot as plt` I thought this was a simple, obvious answer and also opened my eyes to the `seaborn.objects` interface. Thanks! – Tom Bush Mar 21 '23 at 10:20
1

@TomBush i am glad that it helped. added a missing import, thanks! – Maciej Skorski Mar 22 '23 at 07:42
Now we're talking. – Chris Aug 27 '23 at 17:12

score 3 · Answer 5 · answered May 03 '22 at 22:09

Below is a solution that does not iterate over rows in the data frame using the dreaded for loop.

There are many issues regarding iterating over a data frame.

The answer is don't iterate! See this link.

The solution below relies on a function (plotlabel) within the petalplot function, which is called by df.apply.

Now, I know readers will comment on the fact that I use scatter and not lmplot, but that is a bit besides the point.

import pandas as pd
import seaborn as sns
import matplotlib.pyplot as plt
%matplotlib inline

df_iris=sns.load_dataset("iris") 

def petalplot(df): 
    
    def plotlabel(xvar, yvar, label):
        ax.text(xvar+0.002, yvar, label)
        
    fig = plt.figure(figsize=(30,10))
    ax = sns.scatterplot(x = 'sepal_length', y = 'sepal_width', data=df)

    # The magic starts here:
    df.apply(lambda x: plotlabel(x['sepal_length'],  x['sepal_width'], x['species']), axis=1)

    plt.title('Example Plot')
    plt.xlabel('Sepal Length')
    plt.ylabel('Sepal Width')
    
petalplot(df_iris)

The answer that you quote says the following: "You should not use any function with "iter" in its name for more than a few thousand rows or you will have to get used to a lot of waiting." -- It's unlikely that someone would want to put more than that many labels into a plot, so I'd say it's premature optimization from a performance perspective. It's still a pretty elegant solution though. — georch, Aug 14 '22 at 20:14
df.apply is basically a for loop: https://stackoverflow.com/a/52674448/6664393 — user357269, Sep 07 '22 at 21:22

ryu1kn · Answer 6 · 2023-01-03T11:41:48.430

Same idea with Scott Boston's answer, however with Seaborn v0.12+, you can leverage seaborn.FacetGrid.apply to add labels on plots and set up your figure in one go:

import seaborn as sns
import pandas as pd

%matplotlib inline

sns.set_theme()

df_iris = sns.load_dataset("iris")

(
    sns.lmplot(
        data=df_iris,
        x="sepal_length",
        y="sepal_width",
        fit_reg=False,
        height=8,
        aspect=2
    )
    .apply(lambda grid: [
        grid.ax.text(r["sepal_length"]+.02, r["sepal_width"], r["species"])
        for r in df_iris.to_dict(orient="records")
    ])
    .set(title="Example Plot")
    .set_axis_labels("Sepal Length", "Sepal Width")
)

Or, if you don't need to use lmplot, also from v0.12, you can use the seaborn.objects interface. This way we don't need to manually iterate over the Iris dataframe nor refer to df_iris or column names sepal_... multiple times.

import seaborn.objects as so

(
    so.Plot(df_iris, x="sepal_length", y="sepal_width", text="species")
        .add(so.Dot())
        .add(so.Text(halign="left"))
        .label(title="Example plot", x="Sepal Length", y="Sepal Width")
        .layout(size=(20, 10))
)

This produces the below figure:

Adding labels in x y scatter plot with seaborn

6 Answers6

Below is a solution that does not iterate over rows in the data frame using the dreaded for loop.

Linked

Related