binning a dataframe in pandas in Python [duplicate]

There may be a more efficient way (I have a feeling pandas.crosstab would be useful here), but here’s how I’d do it:

import numpy as np
import pandas

df = pandas.DataFrame({"a": np.random.random(100),
                       "b": np.random.random(100),
                       "id": np.arange(100)})

# Bin the data frame by "a" with 10 bins...
bins = np.linspace(df.a.min(), df.a.max(), 10)
groups = df.groupby(np.digitize(df.a, bins))

# Get the mean of each bin:
print groups.mean() # Also could do "groups.aggregate(np.mean)"

# Similarly, the median:
print groups.median()

# Apply some arbitrary function to aggregate binned data
print groups.aggregate(lambda x: np.mean(x[x > 0.5]))

---

Edit: As the OP was asking specifically for just the means of b binned by the values in a, just do

groups.mean().b

Also if you wanted the index to look nicer (e.g. display intervals as the index), as they do in @bdiamante’s example, use pandas.cut instead of numpy.digitize. (Kudos to bidamante. I didn’t realize pandas.cut existed.)

import numpy as np
import pandas

df = pandas.DataFrame({"a": np.random.random(100), 
                       "b": np.random.random(100) + 10})

# Bin the data frame by "a" with 10 bins...
bins = np.linspace(df.a.min(), df.a.max(), 10)
groups = df.groupby(pandas.cut(df.a, bins))

# Get the mean of b, binned by the values in a
print groups.mean().b

This results in:

a
(0.00186, 0.111]    10.421839
(0.111, 0.22]       10.427540
(0.22, 0.33]        10.538932
(0.33, 0.439]       10.445085
(0.439, 0.548]      10.313612
(0.548, 0.658]      10.319387
(0.658, 0.767]      10.367444
(0.767, 0.876]      10.469655
(0.876, 0.986]      10.571008
Name: b