How do I create a multiline plot using seaborn?
I am trying out Seaborn to make my plot visually better than matplotlib. I have a dataset which has a column 'Year' which I want to plot on the X-axis and 4 Columns say A,B,C,D on the Y-axis using different coloured lines. I was trying to do this using the sns.lineplot method but it allows for only one variable on the X-axis and one on the Y-axis. I tried doing this
sns.lineplot(data_preproc['Year'],data_preproc['A'], err_style=None)
sns.lineplot(data_preproc['Year'],data_preproc['B'], err_style=None)
sns.lineplot(data_preproc['Year'],data_preproc['C'], err_style=None)
sns.lineplot(data_preproc['Year'],data_preproc['D'], err_style=None)
But this way I don't get a legend in the plot to show which coloured line corresponds to what. I tried checking the documentation but couldn't find a proper way to do this.
python python-3.x dataframe plot seaborn
asked Sep 13 '18 at 7:35
Shatrunjay PathareShatrunjay Pathare
5615
add a comment |
I am trying out Seaborn to make my plot visually better than matplotlib. I have a dataset which has a column 'Year' which I want to plot on the X-axis and 4 Columns say A,B,C,D on the Y-axis using different coloured lines. I was trying to do this using the sns.lineplot method but it allows for only one variable on the X-axis and one on the Y-axis. I tried doing this
sns.lineplot(data_preproc['Year'],data_preproc['A'], err_style=None)
sns.lineplot(data_preproc['Year'],data_preproc['B'], err_style=None)
sns.lineplot(data_preproc['Year'],data_preproc['C'], err_style=None)
sns.lineplot(data_preproc['Year'],data_preproc['D'], err_style=None)
But this way I don't get a legend in the plot to show which coloured line corresponds to what. I tried checking the documentation but couldn't find a proper way to do this.
python python-3.x dataframe plot seaborn
asked Sep 13 '18 at 7:35
Shatrunjay PathareShatrunjay Pathare
5615
add a comment |
I am trying out Seaborn to make my plot visually better than matplotlib. I have a dataset which has a column 'Year' which I want to plot on the X-axis and 4 Columns say A,B,C,D on the Y-axis using different coloured lines. I was trying to do this using the sns.lineplot method but it allows for only one variable on the X-axis and one on the Y-axis. I tried doing this
sns.lineplot(data_preproc['Year'],data_preproc['A'], err_style=None)
sns.lineplot(data_preproc['Year'],data_preproc['B'], err_style=None)
sns.lineplot(data_preproc['Year'],data_preproc['C'], err_style=None)
sns.lineplot(data_preproc['Year'],data_preproc['D'], err_style=None)
But this way I don't get a legend in the plot to show which coloured line corresponds to what. I tried checking the documentation but couldn't find a proper way to do this.
python python-3.x dataframe plot seaborn
asked Sep 13 '18 at 7:35
Shatrunjay PathareShatrunjay Pathare
5615
I am trying out Seaborn to make my plot visually better than matplotlib. I have a dataset which has a column 'Year' which I want to plot on the X-axis and 4 Columns say A,B,C,D on the Y-axis using different coloured lines. I was trying to do this using the sns.lineplot method but it allows for only one variable on the X-axis and one on the Y-axis. I tried doing this
sns.lineplot(data_preproc['Year'],data_preproc['A'], err_style=None)
sns.lineplot(data_preproc['Year'],data_preproc['B'], err_style=None)
sns.lineplot(data_preproc['Year'],data_preproc['C'], err_style=None)
sns.lineplot(data_preproc['Year'],data_preproc['D'], err_style=None)
But this way I don't get a legend in the plot to show which coloured line corresponds to what. I tried checking the documentation but couldn't find a proper way to do this.
python python-3.x dataframe plot seaborn
python python-3.x dataframe plot seaborn
asked Sep 13 '18 at 7:35
Shatrunjay PathareShatrunjay Pathare
5615
asked Sep 13 '18 at 7:35
Shatrunjay PathareShatrunjay Pathare
5615
asked Sep 13 '18 at 7:35
Shatrunjay PathareShatrunjay Pathare
5615
asked Sep 13 '18 at 7:35
Shatrunjay PathareShatrunjay Pathare
5615
asked Sep 13 '18 at 7:35
Shatrunjay PathareShatrunjay Pathare
5615
5615
add a comment |
add a comment |
2 Answers
2
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oldest
votes
See the documentation:
sns.lineplot(x="Year", y="signal", hue="label", data=data_preproc)
You probably need to re-organize your dataframe in a suitable way so that there is one column for the x data, one for the y data, and one which holds the label for the data point.
You can also just use matplotlib.pyplot. If you import seaborn, much of the improved design is also used for "regular" matplotlib plots. Seaborn is really "just" a collection of methods which conveniently feed data and plot parameters to matplotlib.
answered Sep 13 '18 at 7:42
IonicSolutionsIonicSolutions
865616
add a comment |
This:
sns.lineplot(data=data_preproc)
will do what you want.
answered Nov 24 '18 at 19:55
WolvhLorienWolvhLorien
213
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
See the documentation:
sns.lineplot(x="Year", y="signal", hue="label", data=data_preproc)
You probably need to re-organize your dataframe in a suitable way so that there is one column for the x data, one for the y data, and one which holds the label for the data point.
You can also just use matplotlib.pyplot. If you import seaborn, much of the improved design is also used for "regular" matplotlib plots. Seaborn is really "just" a collection of methods which conveniently feed data and plot parameters to matplotlib.
answered Sep 13 '18 at 7:42
IonicSolutionsIonicSolutions
865616
add a comment |
See the documentation:
sns.lineplot(x="Year", y="signal", hue="label", data=data_preproc)
You probably need to re-organize your dataframe in a suitable way so that there is one column for the x data, one for the y data, and one which holds the label for the data point.
You can also just use matplotlib.pyplot. If you import seaborn, much of the improved design is also used for "regular" matplotlib plots. Seaborn is really "just" a collection of methods which conveniently feed data and plot parameters to matplotlib.
answered Sep 13 '18 at 7:42
IonicSolutionsIonicSolutions
865616
add a comment |
See the documentation:
sns.lineplot(x="Year", y="signal", hue="label", data=data_preproc)
You probably need to re-organize your dataframe in a suitable way so that there is one column for the x data, one for the y data, and one which holds the label for the data point.
You can also just use matplotlib.pyplot. If you import seaborn, much of the improved design is also used for "regular" matplotlib plots. Seaborn is really "just" a collection of methods which conveniently feed data and plot parameters to matplotlib.
answered Sep 13 '18 at 7:42
IonicSolutionsIonicSolutions
865616
See the documentation:
sns.lineplot(x="Year", y="signal", hue="label", data=data_preproc)
You probably need to re-organize your dataframe in a suitable way so that there is one column for the x data, one for the y data, and one which holds the label for the data point.
You can also just use matplotlib.pyplot. If you import seaborn, much of the improved design is also used for "regular" matplotlib plots. Seaborn is really "just" a collection of methods which conveniently feed data and plot parameters to matplotlib.
answered Sep 13 '18 at 7:42
IonicSolutionsIonicSolutions
865616
answered Sep 13 '18 at 7:42
IonicSolutionsIonicSolutions
865616
answered Sep 13 '18 at 7:42
IonicSolutionsIonicSolutions
865616
answered Sep 13 '18 at 7:42
IonicSolutionsIonicSolutions
865616
865616
add a comment |
add a comment |
This:
sns.lineplot(data=data_preproc)
will do what you want.
answered Nov 24 '18 at 19:55
WolvhLorienWolvhLorien
213
add a comment |
This:
sns.lineplot(data=data_preproc)
will do what you want.
answered Nov 24 '18 at 19:55
WolvhLorienWolvhLorien
213
add a comment |
This:
sns.lineplot(data=data_preproc)
will do what you want.
answered Nov 24 '18 at 19:55
WolvhLorienWolvhLorien
213
This:
sns.lineplot(data=data_preproc)
will do what you want.
answered Nov 24 '18 at 19:55
WolvhLorienWolvhLorien
213
edited Nov 24 '18 at 20:08
answered Nov 24 '18 at 19:55
WolvhLorienWolvhLorien
213
answered Nov 24 '18 at 19:55
WolvhLorienWolvhLorien
213
answered Nov 24 '18 at 19:55
WolvhLorienWolvhLorien
213
213
add a comment |
add a comment |
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