Python Numpy 仅按行 shuffle 多维数组，保持列顺序不变
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Numpy shuffle multidimensional array by row only, keep column order unchanged
提问by robert
How can I shuffle a multidimensional array by row only in Python (so do not shuffle the columns).
如何仅在 Python 中按行对多维数组进行混洗（因此不要对列进行混洗）。
I am looking for the most efficient solution, because my matrix is very huge. Is it also possible to do this highly efficient on the original array (to save memory)?
我正在寻找最有效的解决方案，因为我的矩阵非常大。是否也可以在原始数组上高效地执行此操作（以节省内存）？
Example:
例子：
import numpy as np
X = np.random.random((6, 2))
print(X)
Y = ???shuffle by row only not colls???
print(Y)
What I expect now is original matrix:
我现在期望的是原始矩阵：
[[ 0.48252164 0.12013048]
[ 0.77254355 0.74382174]
[ 0.45174186 0.8782033 ]
[ 0.75623083 0.71763107]
[ 0.26809253 0.75144034]
[ 0.23442518 0.39031414]]
Output shuffle the rows not cols e.g.:
输出打乱行而不是列，例如：
[[ 0.45174186 0.8782033 ]
[ 0.48252164 0.12013048]
[ 0.77254355 0.74382174]
[ 0.75623083 0.71763107]
[ 0.23442518 0.39031414]
[ 0.26809253 0.75144034]]
采纳答案by Kasramvd
That's what numpy.random.shuffle()is for :
这numpy.random.shuffle()就是为了：
>>> X = np.random.random((6, 2))
>>> X
array([[ 0.9818058 , 0.67513579],
[ 0.82312674, 0.82768118],
[ 0.29468324, 0.59305925],
[ 0.25731731, 0.16676408],
[ 0.27402974, 0.55215778],
[ 0.44323485, 0.78779887]])
>>> np.random.shuffle(X)
>>> X
array([[ 0.9818058 , 0.67513579],
[ 0.44323485, 0.78779887],
[ 0.82312674, 0.82768118],
[ 0.29468324, 0.59305925],
[ 0.25731731, 0.16676408],
[ 0.27402974, 0.55215778]])
回答by Divakar
You can also use np.random.permutationto generate random permutation of row indices and then index into the rows of Xusing np.takewith axis=0. Also, np.takefacilitates overwriting to the input array Xitself with out=option, which would save us memory. Thus, the implementation would look like this -
您还可以使用np.random.permutation生成行索引的随机排列，然后索引到Xusing np.takewith的行中axis=0。此外，使用选项np.take有助于覆盖输入数组X本身out=，这将节省我们的内存。因此，实现看起来像这样 -
np.take(X,np.random.permutation(X.shape[0]),axis=0,out=X)
Sample run -
样品运行 -
In [23]: X
Out[23]:
array([[ 0.60511059, 0.75001599],
[ 0.30968339, 0.09162172],
[ 0.14673218, 0.09089028],
[ 0.31663128, 0.10000309],
[ 0.0957233 , 0.96210485],
[ 0.56843186, 0.36654023]])
In [24]: np.take(X,np.random.permutation(X.shape[0]),axis=0,out=X);
In [25]: X
Out[25]:
array([[ 0.14673218, 0.09089028],
[ 0.31663128, 0.10000309],
[ 0.30968339, 0.09162172],
[ 0.56843186, 0.36654023],
[ 0.0957233 , 0.96210485],
[ 0.60511059, 0.75001599]])
Additional performance boost
额外的性能提升
Here's a trick to speed up np.random.permutation(X.shape[0])with np.argsort()-
这里有一个窍门，以加快np.random.permutation(X.shape[0])与np.argsort()-
np.random.rand(X.shape[0]).argsort()
Speedup results -
加速结果 -
In [32]: X = np.random.random((6000, 2000))
In [33]: %timeit np.random.permutation(X.shape[0])
1000 loops, best of 3: 510 μs per loop
In [34]: %timeit np.random.rand(X.shape[0]).argsort()
1000 loops, best of 3: 297 μs per loop
Thus, the shuffling solution could be modified to -
因此，洗牌解决方案可以修改为 -
np.take(X,np.random.rand(X.shape[0]).argsort(),axis=0,out=X)
Runtime tests -
运行时测试 -
These tests include the two approaches listed in this post and np.shufflebased one in @Kasramvd's solution.
这些测试包括本文中列出的两种方法并np.shuffle基于@Kasramvd's solution.
In [40]: X = np.random.random((6000, 2000))
In [41]: %timeit np.random.shuffle(X)
10 loops, best of 3: 25.2 ms per loop
In [42]: %timeit np.take(X,np.random.permutation(X.shape[0]),axis=0,out=X)
10 loops, best of 3: 53.3 ms per loop
In [43]: %timeit np.take(X,np.random.rand(X.shape[0]).argsort(),axis=0,out=X)
10 loops, best of 3: 53.2 ms per loop
So, it seems using these np.takebased could be used only if memory is a concern or else np.random.shufflebased solution looks like the way to go.
因此，似乎np.take仅当内存是一个问题时才可以使用这些基于np.random.shuffle的解决方案，否则基于解决方案看起来像是要走的路。
回答by Janmejaya Nanda
After a bit experiment i found most memory and time efficient way to shuffle data(row wise) of nd-array is, shuffle the index and get the data from shuffled index
经过一些实验，我发现对 nd-array 的数据（按行）进行混洗的最有效的内存和时间方法是，对索引进行混洗并从混洗后的索引中获取数据
rand_num2 = np.random.randint(5, size=(6000, 2000))
perm = np.arange(rand_num2.shape[0])
np.random.shuffle(perm)
rand_num2 = rand_num2[perm]
in more details
Here, I am using memory_profilerto find memory usage and python's builtin "time" module to record time and comparing all previous answers
更多细节
在这里，我使用memory_profiler来查找内存使用情况和 python 的内置“时间”模块来记录时间并比较所有以前的答案
def main():
# shuffle data itself
rand_num = np.random.randint(5, size=(6000, 2000))
start = time.time()
np.random.shuffle(rand_num)
print('Time for direct shuffle: {0}'.format((time.time() - start)))
# Shuffle index and get data from shuffled index
rand_num2 = np.random.randint(5, size=(6000, 2000))
start = time.time()
perm = np.arange(rand_num2.shape[0])
np.random.shuffle(perm)
rand_num2 = rand_num2[perm]
print('Time for shuffling index: {0}'.format((time.time() - start)))
# using np.take()
rand_num3 = np.random.randint(5, size=(6000, 2000))
start = time.time()
np.take(rand_num3, np.random.rand(rand_num3.shape[0]).argsort(), axis=0, out=rand_num3)
print("Time taken by np.take, {0}".format((time.time() - start)))
Result for Time
时间结果
Time for direct shuffle: 0.03345608711242676 # 33.4msec
Time for shuffling index: 0.019818782806396484 # 19.8msec
Time taken by np.take, 0.06726956367492676 # 67.2msec
Memory profiler Result
内存分析器结果
Line # Mem usage Increment Line Contents
================================================
39 117.422 MiB 0.000 MiB @profile
40 def main():
41 # shuffle data itself
42 208.977 MiB 91.555 MiB rand_num = np.random.randint(5, size=(6000, 2000))
43 208.977 MiB 0.000 MiB start = time.time()
44 208.977 MiB 0.000 MiB np.random.shuffle(rand_num)
45 208.977 MiB 0.000 MiB print('Time for direct shuffle: {0}'.format((time.time() - start)))
46
47 # Shuffle index and get data from shuffled index
48 300.531 MiB 91.555 MiB rand_num2 = np.random.randint(5, size=(6000, 2000))
49 300.531 MiB 0.000 MiB start = time.time()
50 300.535 MiB 0.004 MiB perm = np.arange(rand_num2.shape[0])
51 300.539 MiB 0.004 MiB np.random.shuffle(perm)
52 300.539 MiB 0.000 MiB rand_num2 = rand_num2[perm]
53 300.539 MiB 0.000 MiB print('Time for shuffling index: {0}'.format((time.time() - start)))
54
55 # using np.take()
56 392.094 MiB 91.555 MiB rand_num3 = np.random.randint(5, size=(6000, 2000))
57 392.094 MiB 0.000 MiB start = time.time()
58 392.242 MiB 0.148 MiB np.take(rand_num3, np.random.rand(rand_num3.shape[0]).argsort(), axis=0, out=rand_num3)
59 392.242 MiB 0.000 MiB print("Time taken by np.take, {0}".format((time.time() - start)))
回答by Ben-Hur Cardoso
You can shuffle a two dimensional array Aby rowusing the np.vectorize()function:
您可以使用以下函数A按行对二维数组进行洗牌np.vectorize()：
shuffle = np.vectorize(np.random.permutation, signature='(n)->(n)')
A_shuffled = shuffle(A)
回答by TassosK
I have a question on this (or maybe it is the answer) Lets say we have a numpy array X with shape=(1000,60,11,1) Also suppose that X is an array of images with size 60x11 and channel number =1 (60x11x1).
我对此有一个问题（或者可能是答案）假设我们有一个形状为 =(1000,60,11,1) 的 numpy 数组 X 还假设 X 是一个大小为 60x11 且通道数 = 的图像数组1 (60x11x1)。
What if I want to shuffle the order of all these images, and to do that I'll use shuffling on the indexes of X.
如果我想打乱所有这些图像的顺序怎么办，为此我将对 X 的索引使用打乱。
def shuffling( X):
indx=np.arange(len(X)) # create a array with indexes for X data
np.random.shuffle(indx)
X=X[indx]
return X
Will that work? From my knowledge len(X) will return the biggest dimension size.
那行得通吗？据我所知 len(X) 将返回最大的尺寸大小。
回答by Ghanem
I tried many solutions, and at the end I used this simple one:
我尝试了很多解决方案，最后我使用了这个简单的解决方案：
from sklearn.utils import shuffle
x = np.array([[1, 2],
[3, 4],
[5, 6]])
print(shuffle(x, random_state=0))
output:
输出：
[
[5 6]
[3 4]
[1 2]
]
if you have 3d array, loop through the 1st axis (axis=0) and apply this function, like:
如果您有 3d 数组，请遍历第一个轴（轴 = 0）并应用此函数，例如：
np.array([shuffle(item) for item in 3D_numpy_array])
