Plotting lists with different number of elements in matplotlib - pandas

Question

I have a list of numpy arrays, each potentially having a different number of elements, such as:
[array([55]),
array([54]),
array([], dtype=float64),
array([48, 55]),]
I would like to plot this, where each array has an abscissa (x value) assigned, such as [1,2,3,4] so that the plot should show the following points: [[1,55], [2, 54], [4, 48], [4, 55]].
Is there a way I can do that with matplotlib? or how can I transform the data with numpy or pandas first so that it is can be plotted?

Answer 1

What you want to do is chain the original array and generate a new array with "abscissas". There are many way to concatenated, one of the most efficient is using itertools.chain.
import itertools
from numpy import array
x = [array([55]), array([54]), array([]), array([48, 55])]
ys = list(itertools.chain(*x))
# this will be [55, 54, 48, 55]
# generate abscissas
xs = list(itertools.chain(*[[i+1]*len(x1) for i, x1 in enumerate(x)]))
Now you can just plot easily with matplotlib as below
import matplotlib.pyplot as plt
plt.plot(xs, ys)

Answer 2

If you want to have different markers for different groups of data (the colours are automatically cycled by matplotlib):
import numpy as np
import matplotlib.pyplot as plt
markers = ['o', #'circle',
'v', #'triangle_down',
'^', #'triangle_up',
'<', #'triangle_left',
'>', #'triangle_right',
'1', #'tri_down',
'2', #'tri_up',
'3', #'tri_left',
'4', #'tri_right',
'8', #'octagon',
's', #'square',
'p', #'pentagon',
'h', #'hexagon1',
'H', #'hexagon2',
'D', #'diamond',
'd', #'thin_diamond'
]
n_markers = len(markers)
a = [10.*np.random.random(int(np.random.random()*10)) for i in xrange(n_markers)]
fig = plt.figure()
ax = fig.add_subplot(111)
for i, data in enumerate(a):
xs = data.shape[0]*[i,] # makes the abscissas list
marker = markers[i % n_markers] # picks a valid marker
ax.plot(xs, data, marker, label='data %d, %s'%(i, marker))
ax.set_xlim(-1, 1.4*len(a))
ax.set_ylim(0, 10)
ax.legend(loc=None)
fig.tight_layout()
Notice the limits to y scale are hard coded, change accordingly. The 1.4*len(a) is meant to leave room on the right side of the graph for the legend.
The example above has no point in the x=0 (would be dark blue circles) as the randomly assigned size for its data set was zero, but you can easily place a +1 if you don't want to use x=0.

Answer 3

Using pandas to create a numpy array with nans inserted when an array is empty or shorter than the longest array in the list...
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
arr_list = [np.array([55]),
np.array([54]),
np.array([], dtype='float64'),
np.array([48, 55]),]
df = pd.DataFrame(arr_list)
list_len = len(df)
repeats = len(list(df))
vals = df.values.flatten()
xax = np.repeat(np.arange(list_len) + 1, repeats)
df_plot = pd.DataFrame({'xax': xax, 'vals': vals})
plt.scatter(df_plot.xax, df_plot.vals);

Answer 4

with x your list :
[plt.plot(np.repeat(i,len(x[i])), x[i],'.') for i in range(len(x))]
plt.show()

Answer 5

#Alessandro Mariani's answer based on itertools made me think of another way to generate an array containg the data I needed. In some cases it may be more compact. It is also based on itertools.chain:
import itertools
from numpy import array
y = [array([55]), array([54]), array([]), array([48, 55])]
x = array([1,2,3,4])
d = array(list(itertools.chain(*[itertools.product([t], n) for t, n in zip(x,y)])))
d is now the following array:
array([[ 1, 55],
[ 2, 54],
[ 4, 48],
[ 4, 55]])