张量上的异或(使用矢量化)在 Tensorflow 中具有浮点值

问题描述

我有两个形状相同的张量 t1 和 t2(在我的例子中是 [64, 64, 3]).我需要计算这两个张量的异或.但想不出办法做到这一点.

I have two tensors t1 and t2 of the same shape (in my case [64, 64, 3]). I need to compute the XOR of these two tensors. But couldn't figure out a way to do so.

import bitstring
from bitstring import *

@tf.function
def xor(x1, x2) :
  a = BitArray(float=x1, length = 64)
  b = BitArray(float=x2, length = 64)
  a ^= b
  return a.float

这个 xor 函数计算 python 中两个浮点值的异或.

This xor function computes xor of two float values in python.

样本输入张量是，

t1 = tf.constant([[1.1, 2.2, 3.3],
                  [4.4, 5.5, 6.6]], dtype=tf.float64)
t2 = tf.constant([[7.7, 8.8, 9.9],
                  [10.1, 11.11, 12.12]], dtype=tf.float64)

我似乎找不到计算两个张量的 xor 的方法.

I can't seem to find a way to compute xor of two tensors.

1. 如何编写矢量化版本的xor 函数调用，该函数调用将从任何形状的两个张量计算每对浮点数的异或(类似于 tf.add、tf.matmul 等)?我试过 np.vectorized 等
2. 我如何高效地编写xor函数?为了在 tensorflow 中使用 GPU，我需要使用 tf.something 编写每个语句，例如tf.add, tf.matmul 等但是由于 tensorflow 没有 Bitstring 的本机支持，有没有办法在 tensorflow(在 xor 函数中)将浮点数转换为比特串，以便我可以稍后再执行 tf.bitwise_xor 吗?

1. How can I write vectorized version of the xor function call which will compute xor of each pair of floats from two tensors of any shape (similar to tf.add, tf.matmul etc)? I tried np.vectorized etc.
2. How can I efficiently write the xor function? In order to use the gpu in tensorflow I need to write each statement using tf.something e.g. tf.add, tf. matmul etc. But since tensorflow doesn't have native support of Bitstring, is there any way to convert float to bitstring in tensorflow (in the xor function) so that I can execute tf.bitwise_xor over that later?

推荐答案

您可能需要自定义 C++ 操作来执行此操作.Tensorflow 文档 有一个关于如何构建一个很好的教程.这是一个让您入门的示例.

You'll probably need a custom C++ op to do this. The Tensorflow docs have a nice tutorial on how to construct one. Here's an example to get you started.

xor_op.cc

#include "tensorflow/core/framework/common_shape_fns.h"
#include "tensorflow/core/framework/op.h"
#include "tensorflow/core/framework/op_kernel.h"
#include "tensorflow/core/framework/shape_inference.h"
#include "tensorflow/core/framework/tensor.h"
#include "tensorflow/core/framework/tensor_types.h"

namespace tensorflow {
using shape_inference::InferenceContext;

REGISTER_OP("Xor")
    .Input("input_tensor_a: float")
    .Input("input_tensor_b: float")
    .Output("output_tensor: float")
    .SetShapeFn([](InferenceContext* c) {
      return shape_inference::UnchangedShapeWithRankAtLeast(c, 1);
    });

class XorOp : public OpKernel {
 public:
  explicit XorOp(OpKernelConstruction* ctx) : OpKernel(ctx) {}

  float XorFloats(const float* a, const float* b, float* c) {
    *(int*)c = *(int*)a ^ *(int*)b;
    return *c;
  }

  void Compute(OpKernelContext* ctx) override {
    // get input tensors
    const Tensor& input_fst = ctx->input(0);
    const Tensor& input_snd = ctx->input(1);

    TTypes<float, 1>::ConstFlat c_in_fst = input_fst.flat<float>();
    TTypes<float, 1>::ConstFlat c_in_snd = input_snd.flat<float>();

    // allocate output tensor
    Tensor* output_tensor = nullptr;
    OP_REQUIRES_OK(ctx,
                   ctx->allocate_output(0, input_fst.shape(), &output_tensor));

    auto output_flat = output_tensor->flat<float>();
    const int N = c_in_fst.size();

    for (int i = 0; i < N; ++i) {
      XorFloats(&c_in_fst(i), &c_in_snd(i), &output_flat(i));
    }
  }
};

REGISTER_KERNEL_BUILDER(Name("Xor").Device(DEVICE_CPU), XorOp);

}  // namespace tensorflow

让我们构建操作并进行测试

Let's build the op and test

$ TF_LFLAGS=($(python -c 'import tensorflow as tf; print(" ".join(tf.sysconfig.get_link_flags()))'))
$ TF_CFLAGS=($(python -c 'import tensorflow as tf; print(" ".join(tf.sysconfig.get_compile_flags()))'))
$
$ g++ -std=c++14 -shared xor_op.cc -o xor_op.so -fPIC ${TF_CFLAGS[@]} ${TF_LFLAGS[@]} -O2

让我们运行 op 看看它是否有效.

Let's run the op and see if it works.

main.py

import tensorflow as tf


def main():
    xor_module = tf.load_op_library("./xor_op.so")
    xor_op = xor_module.xor

    # make some data
    a = tf.constant(
        [[1.1, 2.2, 3.3], [4.4, 5.5, 6.6]],
        dtype=tf.float32)

    b = tf.constant(
        [[7.7, 8.8, 9.9], [10.1, 11.11, 12.12]],
        dtype=tf.float32)

    c = xor_op(a, b)

    print(f"a: {a}")
    print(f"b: {b}")
    print(f"c: {c}")


if __name__ == "__main__":
    main()

# a: [[1.1 2.2 3.3]
#     [4.4 5.5 6.6]]
# b: [[ 7.7   8.8   9.9 ]
#     [10.1  11.11 12.12]]
# c: [[3.3319316e+38 2.3509887e-38 3.7713776e-38]
#     [6.3672620e-38 4.7666294e-38 5.3942895e-38]]

酷.让我们更严格地测试一下.

Cool. Let's test a little more rigorously.

test.py

import tensorflow as tf
from tensorflow.python.platform import test as test_lib


class XorOpTest(test_lib.TestCase):
    def setUp(self):
        # import the custom op
        xor_module = tf.load_op_library("./xor_op.so")
        self._xor_op = xor_module.xor

        # make some data
        self.a = tf.constant(
            [[1.1, 2.2, 3.3], [4.4, 5.5, 6.6]],
            dtype=tf.float32)

        self.b = tf.constant(
            [[7.7, 8.8, 9.9], [10.1, 11.11, 12.12]],
            dtype=tf.float32)

    def test_xor_op(self):
        c = self._xor_op(self.a, self.b)
        self.assertAllEqual(self._xor_op(c, self.b), self.a)


if __name__ == "__main__":
    test_lib.main()

# [ RUN      ] XorOpTest.test_xor_op
# [       OK ] XorOpTest.test_xor_op
# ----------------------------------------------------------------------
# Ran 1 test in 0.005s
#
# OK

我会把它留给你来扩展它以在 GPU 上工作.如果您好奇，XorFloats 方法来自 平方根反问题.

I'll leave it to you to extend this to work on a GPU.If you're curious, the XorFloats method comes from a bit level manipulation used in the inverse square root problem.

这篇关于张量上的异或(使用矢量化)在 Tensorflow 中具有浮点值的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！