I have list of integers and want to get frequency of each integer. This was discussed here
The problem is that approach I\'m using gives me frequency of floating num
(Late to the party, just thought I would add a seaborn
implementation)
seaborn.__version__ = 0.9.0
at time of writing.
Load the libraries and setup mock data.
import numpy as np
import matplotlib.pyplot as plt
import seaborn as sns
data = np.array([3]*10 + [5]*20 + [7]*5 + [9]*27 + [11]*2)
Using specified bins, calculated as per the above question.
sns.distplot(data,bins=np.arange(data.min(), data.max()+1),kde=False,hist_kws={"align" : "left"})
plt.show()
numpy
built-in binning methodsI used the doane
binning method below, which produced integer bins, migth be worth trying out the standard binning methods from numpy.histogram_bin_edges
as this is how matplotlib.hist()
bins the data.
sns.distplot(data,bins="doane",kde=False,hist_kws={"align" : "left"})
plt.show()
Produces the below Histogram:
You can use groupby
from itertools
as shown in How to count the frequency of the elements in a list?
import numpy as np
from itertools import groupby
freq = {key:len(list(group)) for key, group in groupby(np.sort(data))}
If you don't specify what bins to use, np.histogram
and pyplot.hist
will use a default setting, which is to use 10 equal bins. The left border of the 1st bin is the smallest value and the right border of the last bin is the largest.
This is why the bin borders are floating point numbers. You can use the bins
keyword arguments to enforce another choice of bins, e.g.:
plt.hist(data, bins=np.arange(data.min(), data.max()+1))
Edit: the easiest way to shift all bins to the left is probably just to subtract 0.5 from all bin borders:
plt.hist(data, bins=np.arange(data.min(), data.max()+1)-0.5)
Another way to achieve the same effect (not equivalent if non-integers are present):
plt.hist(data, bins=np.arange(data.min(), data.max()+1), align='left')