[FIXED] How to plot list if values with respect to its key of a dictionary in python

Issue

I have a dictionary with list of values

df_param = {};
for i in range(0,1000):
  df_param[i]=[[0]]
print(df_param)

df_param={0: [[0], [20], [20], [20], [5], [1], [5]], 1: [[0], [20], [20], [5], [1], [5]], 2: [[0], [20], [20], [5], [5]], 3: [[0], [20], [5], [5]], 4: [[0], [5], [5]], 5: [[0], [5]], 6: [[0]], 7: [[0]], 8: [[0], [20]], 9: [[0]], 10: [[0]]}

I need to plot each key in x-axis and the list of values in y-axis.

this is my code:

for i in range(0,1000):
  for j in range(0,10):
    curr = df1[j][i]['Classifier'][0]
    if(curr=='KNNClassifier'):
      df_param[i].append([df1[j][i]['Classifier'][1]['n_neighbors']])
print(df_param)

import matplotlib.pyplot as plt

import matplotlib.pyplot as plt
for key, values in df_param.items():
    plt.plot(key, values)
plt.show()

I get the below error.

---------------------------------------------------------------------------
ValueError                                Traceback (most recent call last)
<ipython-input-132-176b8a768873> in <module>
      1 import matplotlib.pyplot as plt
      2 for key, values in df_param.items():
----> 3     plt.plot(key, values)
      4 plt.show()

3 frames
/usr/local/lib/python3.7/dist-packages/matplotlib/axes/_base.py in _plot_args(self, tup, kwargs)
    340 
    341         if x.shape[0] != y.shape[0]:
--> 342             raise ValueError(f"x and y must have same first dimension, but "
    343                              f"have shapes {x.shape} and {y.shape}")
    344         if x.ndim > 2 or y.ndim > 2:

ValueError: x and y must have same first dimension, but have shapes (1,) and (7, 1)

Solution

If your df_param is a dict of form:

{x0: [[y0_a], [y0_b], ...], x1: [[y1_a], [y1_b], ...], ...} and you wish to make a scatter plot of all the (xk, yk_i), then you can first make a proper xy array with two columns x and y:

import numpy as np

xy = np.array([
    (x, y) for x, lst in df_param.items()
    for sublst in lst for y in sublst
])
plt.plot(*xy.T, 'o')

Answered By - Pierre D