Python 计算布尔张量中“真”值的数量
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Count number of "True" values in boolean Tensor
提问by Aidan Gomez
I understand that tf.wherewill return the locations of Truevalues, so that I could use the result's shape[0]to get the number of Trues.
我知道这tf.where将返回True值的位置，以便我可以使用结果shape[0]来获取Trues的数量。
However, when I try and use this, the dimension is unknown (which makes sense as it needs to be computed at runtime). So my question is, how can I access a dimension and use it in an operation like a sum?
但是，当我尝试使用它时，维度是未知的（这是有道理的，因为它需要在运行时计算）。所以我的问题是，我怎样才能访问一个维度并在像求和这样的操作中使用它？
For example:
例如：
myOtherTensor = tf.constant([[True, True], [False, True]])
myTensor = tf.where(myOtherTensor)
myTensor.get_shape() #=> [None, 2]
sum = 0
sum += myTensor.get_shape().as_list()[0] # Well defined at runtime but considered None until then.
采纳答案by Rafa? Józefowicz
You can cast the values to floats and compute the sum on them:
tf.reduce_sum(tf.cast(myOtherTensor, tf.float32))
您可以将值转换为浮点数并计算它们的总和：
tf.reduce_sum(tf.cast(myOtherTensor, tf.float32))
Depending on your actual use case you can also compute sums per row/column if you specify the reduce dimensions of the call.
根据您的实际用例，如果您指定调用的减少维度，您还可以计算每行/列的总和。
回答by mrry
Rafal's answer is almost certainly the simplest way to count the number of trueelements in your tensor, but the other part of your question asked:
Rafal 的答案几乎可以肯定是计算true张量中元素数量的最简单方法，但问题的另一部分提出了：
[H]ow can I access a dimension and use it in an operation like a sum?
[H]如何访问维度并将其用于求和等运算？
To do this, you can use TensorFlow's shape-related operations, which act on the runtime value of the tensor. For example, tf.size(t)produces a scalar Tensorcontaining the number of elements in t, and tf.shape(t)produces a 1D Tensorcontaining the size of tin each dimension.
为此，您可以使用 TensorFlow 的形状相关操作，它们作用于张量的运行时值。例如，tf.size(t)生成Tensor包含 中元素数量的标量t，并tf.shape(t)生成Tensor包含t每个维度中大小的1D 。
Using these operators, your program could also be written as:
使用这些运算符，你的程序也可以写成：
myOtherTensor = tf.constant([[True, True], [False, True]])
myTensor = tf.where(myOtherTensor)
countTrue = tf.shape(myTensor)[0] # Size of `myTensor` in the 0th dimension.
sess = tf.Session()
sum = sess.run(countTrue)
回答by aboettcher
There is a tensorflow function to count non-zero values tf.count_nonzero. The function also accepts an axisand keep_dimsarguments.
有一个 tensorflow 函数来计算非零值tf.count_nonzero。该函数还接受 anaxis和keep_dims参数。
Here is a simple example:
这是一个简单的例子：
import numpy as np
import tensorflow as tf
a = tf.constant(np.random.random(100))
with tf.Session() as sess:
print(sess.run(tf.count_nonzero(tf.greater(a, 0.5))))
回答by Lerner Zhang
I think this is the easiest way to do it:
我认为这是最简单的方法：
In [38]: myOtherTensor = tf.constant([[True, True], [False, True]])
In [39]: if_true = tf.count_nonzero(myOtherTensor)
In [40]: sess.run(if_true)
Out[40]: 3
