The easiest way to plot a pandas dataframe is as described in the documentation like this:
http://pandas.pydata.org/pandas-docs/stable/visualization.html
In my case I want to create a stacked bar chart:
df2.plot(kind='bar', stacked=True);
This is all working well, but I would like to use one column of the df2 as xlabels and not simply have [1,2,3,4... etc] as labels. Is there a simple way to achieve it with an additional parameter in the plot function or do I need to do it in a more complicated way?
The plot uses the index of your dataframe as the labels so if you want to your use use a particular column, set it as your index:
df2.index = df2.labelcol
df2.plot(kind='bar', stacked=True)
Related
I have a dictionary of dataframes where the key is the name of each dataframe and the value is the dataframe itself.
I am looking to iterate through the dictionary and quickly plot the top 10 rows in each dataframe. Each dataframe would have its own plot. I've attempted this with the following:
for df in dfs:
data = dfs[df].head(n=10)
sns.barplot(data=data, x='x_col', y='y_col', color='indigo').set_title(df)
This works, but only returns a plot for the last dataframe in the iteration. Is there a way I can modify this so that I am also able to return the subsequent plots?
By default, seaborn.barplot() plots data on the current Axes. If you didn't specify the Axes to plot on, the latter will override the previous one. To overcome this, you can either create a new figure in each loop or plot on a different axis by specifying the ax argument.
import matplotlib.pyplot as plt
for df in dfs:
data = dfs[df].head(n=10)
plt.figure() # Create a new figure, current axes also changes.
sns.barplot(data=data, x='x_col', y='y_col', color='indigo').set_title(df)
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.
Seaborn favors the "long format" as input. The key ingredient to convert your DataFrame from its "wide format" (one column per measurement type) into long format (one column for all measurement values, one column to indicate the type) is pandas.melt. Given a data_preproc structured like yours, filled with random values:
num_rows = 20
years = list(range(1990, 1990 + num_rows))
data_preproc = pd.DataFrame({
'Year': years,
'A': np.random.randn(num_rows).cumsum(),
'B': np.random.randn(num_rows).cumsum(),
'C': np.random.randn(num_rows).cumsum(),
'D': np.random.randn(num_rows).cumsum()})
A single plot with four lines, one per measurement type, is obtained with
sns.lineplot(x='Year', y='value', hue='variable',
data=pd.melt(data_preproc, ['Year']))
(Note that 'value' and 'variable' are the default column names returned by melt, and can be adapted to your liking.)
This:
sns.lineplot(data=data_preproc)
will do what you want.
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.
I want I stacked histogram where the different classes are visible.
At the moment I have the histogram without classes with this code:
plt.hist(hist_matrix2.column_name)
which produces this histogram:
and another histogram with the same data, that is grouped by the classes with this code:
hist_matrix2.groupby("number").column_name.plot.hist(alpha=0.5, bins = [0,5,10,15,20,25,30], stacked = True)
which produces this histogram:
As you can see the classes are there but it is not stacked, although the parameter is set. What can I do to stack the classes?
plt.hist has a built-in stacking flag you can set:
plt.hist(hist_matrix2.column_name, stacked=True)
Edit in response to your question, for long data (with multiple levels stacked) first you need to restructure the data into a list of lists:
wide=hist_matrix2.pivot( columns='number', values='column_name')
#This creates many missing values which pandas does not like, so we drop them
widelist=[wide[col].dropna() for col in wide.columns]
# and the stacked graph is here
plt.hist(widelist,stacked=True)
plt.show()
I have a dataset of 5000 products with 50 features. One of the column is 'colors' and there are more than 100 colors in the column. I'm trying to plot a bar chart to show only the top 10 colors and how many products there are in each color.
top_colors = df.colors.value_counts()
top_colors[:10].plot(kind='barh')
plt.xlabel('No. of Products');
Using Seaborn:
sns.factorplot("colors", data=df , palette="PuBu_d");
1) Is there a better way to do this?
2) How can i replicate this with Seaborn?
3) How do i plot such that the highest count is at the top (i.e black at the very top of the bar chart)
An easy trick might be to invert the y axis of your plot, rather than futzing with the data:
s = pd.Series(np.random.choice(list(string.uppercase), 1000))
counts = s.value_counts()
ax = counts.iloc[:10].plot(kind="barh")
ax.invert_yaxis()
Seaborn barplot doesn't currently support horizontally oriented bars, but if you want to control the order the bars appear in you can pass a list of values to the x_order param. But I think it's easier to use the pandas plotting methods here, anyway.
If you want to use pandas then you can first sort:
top_colors[:10].sort(ascending=0).plot(kind='barh')
Seaborn already styles your pandas plots, but you can also use:
sns.barplot(top_colors.index, top_colors.values)
I have time series in a Pandas dateframe with a number of columns which I'd like to plot. Is there a way to set the x-axis to always use the index from a dateframe?
When I use the .plot() method from Pandas the x-axis is formatted correctly however I when I pass my dates and the column(s) I'd like to plot directly to matplotlib the graph doesn't plot correctly. Thanks in advance.
plt.plot(site2.index.values, site2['Cl'])
plt.show()
FYI: site2.index.values produces this (I've cut out the middle part for brevity):
array([
'1987-07-25T12:30:00.000000000+0200',
'1987-07-25T16:30:00.000000000+0200',
'2010-08-13T02:00:00.000000000+0200',
'2010-08-31T02:00:00.000000000+0200',
'2010-09-15T02:00:00.000000000+0200'
],
dtype='datetime64[ns]')
It seems the issue was that I had .values. Without it (i.e. site2.index) the graph displays correctly.
You can use plt.xticks to set the x-axis
try:
plt.xticks( site2['Cl'], site2.index.values ) # location, labels
plt.plot( site2['Cl'] )
plt.show()
see the documentation for more details: http://matplotlib.org/api/pyplot_api.html#matplotlib.pyplot.xticks
That's Builtin Right Into To plot() method
You can use yourDataFrame.plot(use_index=True) to use the DataFrame Index On X-Axis.
The "use_index=True" sets the DataFrame Index on the X-Axis.
Read More Here: https://pandas.pydata.org/pandas-docs/version/0.23/generated/pandas.DataFrame.plot.html
you want to use matplotlib to select a 'sensible' scale just like me, there is one way can solve this question. using a Pandas dataframe index as values for x-axis in matplotlib plot. Code:
ax = plt.plot(site2['Cl'])
x_ticks = ax.get_xticks() # use matplotlib default xticks
x_ticks = list(filter(lambda x: x in range(len(site2)), x_ticks))
ax.set_xticklabels([' '] + site2.index.iloc[x_ticks].to_list())