问题

我正在尝试找到TensorFlow中与np.quantile()1等效的方法。我找到了tfp.stats.quantiles()2（其中tfp代表TensorFlow Probability）。然而，它的构造与np.quantile()有一些不同。

考虑以下示例：

import tensorflow_probability as tfp
import tensorflow as tf
import numpy as np

inputs = tf.random.normal((1, 4096, 4))

print("NumPy")
print(np.quantile(inputs.numpy(), q=0.9, axis=1, keepdims=False))

我不确定如何使用tfp.stats.quantile()来实现上述内容，从TFP文档中也无法确定。我尝试查看了两种方法的源代码，但没有帮助。

import tensorflow_probability as tfp
import tensorflow as tf 
import numpy as np 

inputs = tf.random.normal((1, 4096, 4))

print("NumPy")
print(np.quantile(inputs.numpy(), q=0.9, axis=1, keepdims=False))

答案1

得分: 2

以下是翻译好的部分：

让我在这里试着更有帮助些，而不像我之前在 GitHub 上的表现。

`np.quantile` 和 `tfp.stats.quantiles` 之间存在行为差异。关键的区别在于 `numpy.quantile` 会

> 计算沿指定轴的 q 分位数。

其中 `q` 是

> 要计算的分位数或分位数序列，必须介于 0 和 1 之间（包括0和1）。

而 `tfp.stats.quantiles`

> 给定样本向量 `x`，此函数通过返回 `num_quantiles + 1` 个分位点来估算分位点

因此，您需要告诉 `tfp.stats.quantiles` 您想要多少个分位数，然后选择 `q` 分位数。如果从 API 中不清楚如何做到这一点，如果您查看 `tfp.stats.quantiles` 的[源代码](https://github.com/tensorflow/probability/blob/0759c57eb0306a5bb0fcc3d105bbcab9943092a5/tensorflow_probability/python/stats/quantiles.py#L727-L741)（对于 `v0.19.0`），我们可以看到它显示了如何获得与 NumPy 相似的返回结构。

为了完整起见，使用以下方式设置虚拟环境：

```console
$ cat requirements.txt
numpy==1.24.2
tensorflow==2.11.0
tensorflow-probability==0.19.0

使我们能够运行

import numpy as np
import tensorflow as tf
import tensorflow_probability as tfp

inputs = tf.random.normal((1, 4096, 4), dtype=tf.float64)
q = 0.9

numpy_quantiles = np.quantile(inputs.numpy(), q=q, axis=1, keepdims=False)

tfp_quantiles = tfp.stats.quantiles(
    inputs, num_quantiles=100, axis=1, interpolation="linear"
)[int(q * 100)]

assert np.allclose(numpy_quantiles, tfp_quantiles.numpy())

print(f"{numpy_quantiles=}")
# numpy_quantiles=array([[1.31727661, 1.2699167 , 1.28735237, 1.27137588]])
print(f"{tfp_quantiles=}")
# tfp_quantiles=<tf.Tensor: shape=(1, 4), dtype=float64, numpy=array([[1.31727661, 1.2699167 , 1.28735237, 1.27137588])>

<details>
<summary>英文:</summary>

Let me try to be more helpful here than [I was on GitHub](https://github.com/tensorflow/probability/issues/864#issuecomment-1416956051).

There is a difference in behavior between `np.quantile` and `tfp.stats.quantiles`. The key difference here is that `numpy.quantile` will

&gt; Compute the q-th quantile of the data along the specified axis.

where `q` is the

&gt; Quantile or sequence of quantiles to compute, which must be between 0 and 1 inclusive.

and `tfp.stats.quantiles`

&gt; Given a vector `x` of samples, this function estimates the cut points by returning `num_quantiles + 1` cut points

So you need to tell `tfp.stats.quantiles` how many quantiles you want and then select out the `q`th quantile. If it isn&#39;t clear how to do this just from the API, if you look at the [source for `tfp.stats.quantiles`](https://github.com/tensorflow/probability/blob/0759c57eb0306a5bb0fcc3d105bbcab9943092a5/tensorflow_probability/python/stats/quantiles.py#L727-L741) (for `v0.19.0`) we can see that it shows us how we can get a similar return structure as NumPy.

For completeness, setting up a virtual environment with

```console
$ cat requirements.txt
numpy==1.24.2
tensorflow==2.11.0
tensorflow-probability==0.19.0

allows us to run

import numpy as np
import tensorflow as tf
import tensorflow_probability as tfp

inputs = tf.random.normal((1, 4096, 4), dtype=tf.float64)
q = 0.9

numpy_quantiles = np.quantile(inputs.numpy(), q=q, axis=1, keepdims=False)

tfp_quantiles = tfp.stats.quantiles(
    inputs, num_quantiles=100, axis=1, interpolation="linear"
)[int(q * 100)]

assert np.allclose(numpy_quantiles, tfp_quantiles.numpy())

print(f"{numpy_quantiles=}")
# numpy_quantiles=array([[1.31727661, 1.2699167 , 1.28735237, 1.27137588]])
print(f"{tfp_quantiles=}")
# tfp_quantiles=<tf.Tensor: shape=(1, 4), dtype=float64, numpy=array([[1.31727661, 1.2699167 , 1.28735237, 1.27137588]])>

答案2

得分: 1

你也可以使用 tfp.stats.percentile(inputs, 90., axis=1, keepdims=False) -- 唯一的区别是 90. 替代了 .90.