如何使用Pandas创建随机整数的DataFrame? [英] How to create a DataFrame of random integers with Pandas?
问题描述
我知道,如果我使用 randn
,
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'))
给我想要的东西,但带有正态分布的元素.但是,如果我只想要随机整数怎么办?
gives me what I am looking for, but with elements from a normal distribution. But what if I just wanted random integers?
randint
通过提供范围来工作,但不能像randn
那样提供数组.那么我该如何使用某个范围之间的随机整数呢?
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?
推荐答案
numpy.random.randint
接受第三个参数(size
),您可以在其中指定输出数组的大小.您可以使用它来创建您的DataFrame
-
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'))
在这里-np.random.randint(0,100,size=(100, 4))
-创建大小为(100,4)
的输出数组,并在[0,100)
之间插入随机整数元素.
Here - np.random.randint(0,100,size=(100, 4))
- creates an output array of size (100,4)
with random integer elements between [0,100)
.
演示-
import numpy as np
import pandas as pd
df = pd.DataFrame(np.random.randint(0,100,size=(100, 4)), columns=list('ABCD'))
产生:
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
.. .. .. .. ..
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