Python reduce_sum() 在 tensorflow 中是如何工作的？
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How does reduce_sum() work in tensorflow?
提问by Bhaskar Dhariyal
I am learning tensorflow, I picked up the following code from the tensorflow website. According to my understanding, axis=0 is for rows and axis=1 is for columns.
我正在学习 tensorflow，我从 tensorflow 网站上获取了以下代码。根据我的理解，axis=0 表示行，axis=1 表示列。
How are they getting output mentioned in comments? I have mentioned output according to my thinking against ##.
他们如何获得评论中提到的输出？我已经根据我对## 的想法提到了输出。
import tensorflow as tf
x = tf.constant([[1, 1, 1], [1, 1, 1]])
tf.reduce_sum(x, 0) # [2, 2, 2] ## [3, 3]
tf.reduce_sum(x, 1) # [3, 3] ##[2, 2, 2]
tf.reduce_sum(x, [0, 1]) # 6 ## Didn't understood at all.
回答by Dmitriy Danevskiy
xhas a shape of (2, 3)(two rows and three columns):
x具有(2, 3)（两行三列）的形状：
1 1 1
1 1 1
By doing tf.reduce_sum(x, 0)the tensor is reduced along the first dimension (rows), so the result is [1, 1, 1] + [1, 1, 1] = [2, 2, 2].
通过这样做tf.reduce_sum(x, 0)，张量沿着第一维（行）减少，所以结果是[1, 1, 1] + [1, 1, 1] = [2, 2, 2]。
By doing tf.reduce_sum(x, 1)the tensor is reduced along the second dimension (columns), so the result is [1, 1] + [1, 1] + [1, 1] = [3, 3].
通过这样做tf.reduce_sum(x, 1)，张量沿着第二个维度（列）减少，所以结果是[1, 1] + [1, 1] + [1, 1] = [3, 3]。
By doing tf.reduce_sum(x, [0, 1])the tensor is reduced along BOTH dimensions (rows and columns), so the result is 1 + 1 + 1 + 1 + 1 + 1 = 6or, equivalently, [1, 1, 1] + [1, 1, 1] = [2, 2, 2], and then 2 + 2 + 2 = 6(reduce along rows, then reduce the resulted array).
通过这样做tf.reduce_sum(x, [0, 1])，张量沿两个维度（行和列）减少，因此结果是1 + 1 + 1 + 1 + 1 + 1 = 6或等价地，[1, 1, 1] + [1, 1, 1] = [2, 2, 2]，然后2 + 2 + 2 = 6（沿行减少，然后减少结果数组）。
回答by Maxim
The input is a 2-D tensor:
输入是一个二维张量：
1 1 1
1 1 1
The 0 axis in tensorflow is the rows, 1 axis is the columns. The sum along the 0 axiswill produce a 1-D tensor of length 3, each element is a per-column sum. The result is thus [2, 2, 2]. Likewise for the rows.
tensorflow 中的 0 轴是行，1 轴是列。沿 0 轴的总和将产生一个长度为 的一维张量3，每个元素是每列的总和。结果是这样的[2, 2, 2]。行也是如此。
The sum along both axes is, in this case, the sum of all values in the tensor, which is 6.
在这种情况下，沿两个轴的总和是张量中所有值的总和，即6。
Comparison to numpy:
与numpy 的比较：
a = np.array([[1, 1, 1], [1, 1, 1]])
np.sum(a, axis=0) # [2 2 2]
np.sum(a, axis=1) # [3 3]
np.sum(a, axis=(0, 1)) # 6
As you can see, the output is the same.
如您所见，输出是相同的。
回答by Enrique Ortu?o
In order to understand better what is going on I will change the values, and the results are self explanatory
为了更好地了解发生了什么，我将更改值，结果不言自明
import tensorflow as tf
x = tf.constant([[1, 2, 4], [8, 16, 32]])
a = tf.reduce_sum(x, 0) # [ 9 18 36]
b = tf.reduce_sum(x, 1) # [ 7 56]
c = tf.reduce_sum(x, [0, 1]) # 63
with tf.Session() as sess:
output_a = sess.run(a)
print(output_a)
output_b = sess.run(b)
print(output_b)
output_c = sess.run(c)
print(output_c)
回答by asakryukin
Think it like that, the axis indicates the dimension which will be eliminated. So for the first case axis 0, so if you go through this dimension (2 entries) they will all collapse into 1. Thus it will be as following:
这样想，轴表示将被淘汰的维度。所以对于第一个 case axis 0，所以如果你通过这个维度（2 个条目），它们将全部折叠为 1。因此它将如下所示：
result = [[1,1,1] + [1,1,1]] = [2,2,2]
So you removed dimension 0.
所以你删除了维度0。
Now, for the second case, you will collapse axis 1(or columns), so:
现在，对于第二种情况，您将折叠轴1（或列），因此：
result = [[1,1] + [1,1] + [1,1]] = [3,3]
And the last case is you keep collapsing in order indicated in the brackets. In other words, first you eliminate the rows and then the columns:
最后一种情况是您继续按括号中指示的顺序折叠。换句话说，首先消除行，然后消除列：
result1 = [2,2,2]
result_final = 2 + 2 + 2 = 6
Hope this helps!
希望这可以帮助！
回答by Ashiq Imran
tf.reduce_sum(x, [0, 1])
commands will calculate sum across axis = 0 (row-wise) first, then will calculate sum across axis = 1 (column-wise)
命令将首先计算跨轴的总和 = 0（按行），然后计算跨轴的总和 = 1（按列）
For example,
例如，
x = tf.constant([[1, 1, 1], [1, 1, 1]])
You are summing into [2,2,2] after calculating sum across axis = 0. You are summing 2 + 2 + 2 after calculating sum across axis = 1.
在计算跨轴总和 = 0 后求和为 [2,2,2]。在计算跨轴总和 = 1 后求和 2 + 2 + 2。
Finally, getting 6 as output.
最后，得到 6 作为输出。
