Is there a parameter in matplotlib/pandas to have the Y axis of a histogram as percentage?

The density=True (normed=True for matplotlib < 2.2.0) returns a histogram for which np.sum(pdf * np.diff(bins)) equals 1. If you want the sum of the histogram to be 1 you can use Numpy's histogram() and normalize the results yourself.

x = np.random.randn(30)

fig, ax = plt.subplots(1,2, figsize=(10,4))

ax[0].hist(x, density=True, color='grey')

hist, bins = np.histogram(x)
ax[1].bar(bins[:-1], hist.astype(np.float32) / hist.sum(), width=(bins[1]-bins[0]), color='grey')

ax[0].set_title('normed=True')
ax[1].set_title('hist = hist / hist.sum()')

Btw: Strange plotting glitch at the first bin of the left plot.

Pandas plotting can accept any extra keyword arguments from the respective matplotlib function. So for completeness from the comments of others here, this is how one would do it:

import pandas as pd
import numpy as np

df = pd.DataFrame(np.random.randn(100,2), columns=list('AB'))

df.hist(density=1)

Also, for direct comparison this may be a good way as well:

df.plot(kind='hist', density=1, bins=20, stacked=False, alpha=.5)

Looks like @CarstenKönig found the right way:

df.hist(bins=20, weights=np.ones_like(df[df.columns[0]]) * 100. / len(df))

I know this answer is 6 years later but to anyone using density=True (the substitute for the normed=True), this is not doing what you might want to. It will normalize the whole distribution so that the area of the bins is 1. So if you have more bins with a width < 1 you can expect the height to be > 1 (y-axis). If you want to bound your histogram to [0;1] you will have to calculate it yourself.