Color matplotlib bar chart based on value

Question

Is there a way to color the bars of a barchart based on the bar's value. For example:

- values below -0.5: red
- values between -0.5 to 0: green
- values between 0 to 08: blue
- etc

I have found some basic examples of bar coloring but nothing which can cater for value ranges, such as the above examples.

UPDATE:

Thank you kikocorreoso for your suggestion. This works great when both axes are numbers as per your example. However in my case my original data structure is a pandas dataframe. I then use df.stack() and plot the result. This means that the dataframes rows/columns become the x axis of the plot and the dataframe cells are the Y axis (bars).

I have tried masking as per your example but it doesn't seem to work when the Y axis are numbers and the X axis are names. eg:

     col1    col2   col3   col4
 row1 1       2      3      4
 row2 5       6      7      8
 row3 9       10     11     12
 row4 13      14     15     16

The above dataframe needs to be plotted as a barchart with the row/column combinations forming the x-axis. Each cell value will be a bar. And ultimately, coloring the bars as per the original question. Thanks

Answer 1

With list comprehension and conditional expressions:

profit_color = [('green' if p > 0 else 'red') for p in profit]

plt.bar(year, profit, color=profit_color)

Full example with value ranges:

import matplotlib.pyplot as plt

profit = [-4.56, -0.81, -0.47, 0.25, 1.05, 1.93, 2.91, 3.43]
year = [*range(len(profit))]

profit_color = [{p<0: 'red', 0<=p<=2: 'orange', p>2: 'green'}[True] for p in profit]

plt.bar(year, profit, color=profit_color)
plt.show()

Output:

Answer 2

small improvement to the answer above

x = np.arange(4)

y = np.array([-2,3,-1,2])

cc=list(map(lambda x: 'red' if x <= 0 else 'blue', y))

plt.bar(x, y, color = cc)
plt.show()

Answer 3

import numpy as np
import matplotlib.pyplot as plt

x = np.arange(4)
y = np.array([-2,3,-1,2])
cc=['colors']*len(y)
for n,val in enumerate(y):
    if val<0:
        cc[n]='red'
    elif val>=0
        cc[n]='blue'

plt.bar(x, y, color = cc)
plt.show()

Answer 4

I see this question has been asked a long time ago already, but just in case it could help someone, this is what worked for me:

Iterate over all values and append colors to a list depending on customized conditions, so you get a list with as many color specifications as you have values; then use the color list in plt.bar():

import numpy as np
import matplotlib.pyplot as plt

x = np.arange(10)
y = np.arange(10) * 0.1

col = []
for val in y:
    if val < 0.4:
        col.append('blue')
    elif val >= 0.7:
        col.append('green')
    else:
        col.append('red')

# col looks like this: ['blue', 'blue', 'blue', 'blue', 'red', 'red', 'red', 'green', 'green', 'green']

plt.bar(x, y, color = col)

Answer 5

You could use masks for your datasets. A basic example could be the following:

import numpy as np
import matplotlib.pyplot as plt

x = np.arange(10)
y = np.arange(10) * 0.1

mask1 = y < 0.5
mask2 = y >= 0.5

plt.bar(x[mask1], y[mask1], color = 'red')
plt.bar(x[mask2], y[mask2], color = 'blue')
plt.show()

The result should be:

UPDATE:

As you updated your question I update the code. For your simple case, and if I understood correctly, you could do the following (ugly) hack:

import pandas as pd

df = pd.DataFrame({'col1':[1,2,3], 'col2':[4,5,6]}, 
                  index = ['row1','row2','row3'])

dfstacked = df.stack()

mask = dfstacked <= 3

colors = np.array(['b']*len(dfstacked))
colors[mask.values] = 'r'

dfstacked.plot(kind = 'bar', rot = 45, color = colors)
plt.show()

Or use a more OO solution.

The code briefly explained:

• I create a mask for my red columns
• I create an array of colors
• Change the the array of colors in order to use other color for my masked values
• As the dfstacked dataframe has a MultiIndex the ticks are not well printed so I use the rot keyword to rotate them. If you want to automate it in order to get a nice plot you can use plt.tight_layout() before plt.show().

I hope it helps.