# How to create a DataFrame of random integers with Pandas?

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### Question :

How to create a DataFrame of random integers with Pandas?

I know that if I use `randn`,

``````import pandas as pd
import numpy as np
df = pd.DataFrame(np.random.randn(100, 4), columns=list('ABCD'))
``````

gives me what I am looking for, but with elements from a normal distribution. But what if I just wanted random integers?

`randint` works by providing a range, but not an array like `randn` does. So how do I do this with random integers between some range?

`numpy.random.randint` accepts a third argument (`size`) , in which you can specify the size of the output array. You can use this to create your `DataFrame`

``````df = pd.DataFrame(np.random.randint(0,100,size=(100, 4)), columns=list('ABCD'))
``````

Here – `np.random.randint(0,100,size=(100, 4))` – creates an output array of size `(100,4)` with random integer elements between `[0,100)` .

Demo –

``````import numpy as np
import pandas as pd
df = pd.DataFrame(np.random.randint(0,100,size=(100, 4)), columns=list('ABCD'))
``````

which produces:

``````     A   B   C   D
0   45  88  44  92
1   62  34   2  86
2   85  65  11  31
3   74  43  42  56
4   90  38  34  93
5    0  94  45  10
6   58  23  23  60
..  ..  ..  ..  ..
``````

The recommended way to create random integers with NumPy these days is to use `numpy.random.Generator.integers`. (documentation)

``````import numpy as np
import pandas as pd

rng = np.random.default_rng()
df = pd.DataFrame(rng.integers(0, 100, size=(100, 4)), columns=list('ABCD'))
df
----------------------
A    B    C    D
0   58   96   82   24
1   21    3   35   36
2   67   79   22   78
3   81   65   77   94
4   73    6   70   96
... ...  ...  ...  ...
95   76   32   28   51
96   33   68   54   77
97   76   43   57   43
98   34   64   12   57
99   81   77   32   50
100 rows × 4 columns
``````