How can I shuffle a multidimensional array by row only in Python (so do not shuffle the columns).
我怎样才能在Python中按行对多维数组进行洗牌(所以不要随意乱洗列)。
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,因为我的矩阵非常庞大。是否也可以在原始阵列上高效地工作(以节省内存)?
Example:
例:
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.:
输出洗牌行没有cols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]]
3 个解决方案
#1
16
That's what numpy.random.shuffle()
is for :
这就是numpy.random.shuffle()的用途:
>>> 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]])
#2
11
You can also use np.random.permutation
to generate random permutation of row indices and then index into the rows of X
using np.take
with axis=0
. Also, np.take
facilitates overwriting to the input array X
itself 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,然后使用轴= 0的np.take将索引转换为X行。此外,np.take有助于使用out =选项覆盖输入数组X本身,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()
-
这是使用np.argsort()加速np.random.permutation(X.shape [0])的技巧 -
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,改组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.shuffle
based one in @Kasramvd's solution
.
这些测试包括本文中列出的两种方法和基于@ Kasramvd解决方案的np.shuffle。
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.take
based could be used only if memory is a concern or else np.random.shuffle
based solution looks like the way to go.
因此,似乎使用这些基于np.take的方法只有在内存成为问题时才能使用,否则基于np.random.shuffle的解决方案就像是要走的路。
#3
2
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,我发现大多数内存和时间有效的方式来重新排列nd-array的数据(行方式),将索引洗牌并从混洗索引中获取数据
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_profiler to 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
时间的结果
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)))
#1
16
That's what numpy.random.shuffle()
is for :
这就是numpy.random.shuffle()的用途:
>>> 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]])
#2
11
You can also use np.random.permutation
to generate random permutation of row indices and then index into the rows of X
using np.take
with axis=0
. Also, np.take
facilitates overwriting to the input array X
itself 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,然后使用轴= 0的np.take将索引转换为X行。此外,np.take有助于使用out =选项覆盖输入数组X本身,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()
-
这是使用np.argsort()加速np.random.permutation(X.shape [0])的技巧 -
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,改组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.shuffle
based one in @Kasramvd's solution
.
这些测试包括本文中列出的两种方法和基于@ Kasramvd解决方案的np.shuffle。
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.take
based could be used only if memory is a concern or else np.random.shuffle
based solution looks like the way to go.
因此,似乎使用这些基于np.take的方法只有在内存成为问题时才能使用,否则基于np.random.shuffle的解决方案就像是要走的路。
#3
2
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,我发现大多数内存和时间有效的方式来重新排列nd-array的数据(行方式),将索引洗牌并从混洗索引中获取数据
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_profiler to 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
时间的结果
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)))