12.3.10.4.3. Grouping data

import pandas as pd
import numpy as np

Concat

Create a dataFrame

dataFrame = pd.DataFrame(np.random.randn(10, 4))

break in pieces

pieces = [dataFrame[:3], dataFrame[3:7], dataFrame[7:]]

pd.concat(pieces)

	0	1	2	3
0	2.005605	0.122321	0.314603	0.318486
1	-1.015193	0.074791	0.816450	-1.115685
2	-0.997419	-0.064948	0.973292	0.969946
3	1.281346	0.154516	-3.354396	-0.940462
4	-0.063201	0.225435	0.350171	1.603465
5	0.654310	0.085126	0.021388	-0.631416
6	1.328035	0.053975	0.672767	-1.708392
7	-0.512168	-0.044290	0.993882	-0.127941
8	1.000896	-0.096002	-0.327849	2.242147
9	0.407832	-0.135254	1.369629	0.255078

Join

left = pd.DataFrame({"key": ["foo", "foo"], "lval": [1, 2]})
right = pd.DataFrame({"key": ["foo", "foo"], "rval": [4, 5]})

pd.merge(left, right, on="key")

	key	lval	rval
0	foo	1	4
1	foo	1	5
2	foo	2	4
3	foo	2	5

Grouping

dataFrame = pd.DataFrame(
    {
        "A": ["foo", "bar", "foo", "bar", "foo", "bar", "foo", "foo"],
        "B": ["one", "one", "two", "three", "two", "two", "one", "three"],
        "C": np.random.randn(8),
        "D": np.random.randn(8),
    }
)
dataFrame.groupby("A").sum()

	B	C	D
A
bar	onethreetwo	2.097917	0.999664
foo	onetwotwoonethree	2.141605	0.976511

dataFrame.groupby(["A", "B"]).sum()

		C	D
A	B
bar	one	0.048469	0.013294
	three	0.631859	-0.183019
	two	1.417589	1.169390
foo	one	1.114664	1.950852
	three	-0.457640	-2.412908
	two	1.484580	1.438567