Setting colors individually in matplotlib - numpy

Question

I want to create a custom plot. I want to precisely specify the color of each object. Specifically, I am creating a Gantt chart for system events. I am classifying those events into groups and color coding them to visualize.
Please consider the following code:
import matplotlib.pyplot as plt
import pandas as pd
df = pd.DataFrame()
df['y'] = [0,4,5,6,10]
df['color'] = [(.5, .5, .5, .5),]*len(df)
print df['color']
#fig = plt.figure(figsize=(12, 6))
#vax = fig.add_subplot(1,1,1)
#vax.hlines(df['y'], 0, 10, colors=df['color'])
#fig.savefig('ok.png')
only_four = df['y']==4
df['color'][only_four] = [(0.7, 0.6, 0.5, 0.4),]*sum(only_four)
print df['color']
Note that I first am setting the color for all to be a semi-transparent gray. Later, for a particular set of values, I want to change the color. I end up with this color table.
0 (0.5, 0.5, 0.5, 0.5)
1 0.6
2 (0.5, 0.5, 0.5, 0.5)
3 (0.5, 0.5, 0.5, 0.5)
4 (0.5, 0.5, 0.5, 0.5)
I want to be able to specify any RGBA value (i.e. including transparency) for any subset of the hlines. Could someone share how to do this? I'm open to any other way to do this as long as I can precisely color each line including a transparency.
ADDITION TO QUESTION:
I am able to update multiple rows by iterating as in:
def set_color(df, row_bool, r, g, b, a=1.0):
idx = np.where(row_bool)[0]
for i in idx:
df['color'][i] = (r,g,b,a)
return
This is sufficient, but I really wanted a vector operation (ie no explicit loop by me).

Answer 1

I'm guessing the problem is that you cannot get your updated tuple to be input into the DataFrame and you only get that 0.6 value in the DataFrame. Have you tried using DataFrame.set_value?
In [1]: df
Out[1]:
y color
0 0 (0.5, 0.5, 0.5, 0.5)
1 4 0.6
2 5 (0.5, 0.5, 0.5, 0.5)
3 6 (0.5, 0.5, 0.5, 0.5)
4 10 (0.5, 0.5, 0.5, 0.5)
In [2]: df.set_value(1, 'color', (0.7, 0.6, 0.5, 0.4))
Out[2]:
y color
0 0 (0.5, 0.5, 0.5, 0.5)
1 4 (0.7, 0.6, 0.5, 0.4)
2 5 (0.5, 0.5, 0.5, 0.5)
3 6 (0.5, 0.5, 0.5, 0.5)
4 10 (0.5, 0.5, 0.5, 0.5)