TF常用知识

命名空间及变量共享

# coding=utf-8
import tensorflow as tf
import numpy as np
import matplotlib.pyplot as plt;

with tf.variable_scope('V1') as scope:
    a1 = tf.get_variable(name='a1', shape=[1], initializer=tf.constant_initializer(1))
    scope.reuse_variables()
    a3 = tf.get_variable('a1')

with tf.Session() as sess:
    sess.run(tf.initialize_all_variables())
    print a1.name
    print sess.run(a1)
    print a3.name
    print sess.run(a3)

等同于


with tf.variable_scope('V1'):
	a1 = tf.get_variable(name='a1', shape=[1], initializer=tf.constant_initializer(1))
 
with tf.variable_scope('V1', reuse=True):
	a3 = tf.get_variable('a1')
 
with tf.Session() as sess:
	sess.run(tf.initialize_all_variables())
	print a1.name
	print sess.run(a1)
	print a3.name
	print sess.run(a3)

数学运算

基本数学函数

相同大小Tensor之间的任何算术运算都会将运算应用到元素级。

# 算术操作符:+ - * / % 
tf.add(x, y, name=None)        # 加法(支持 broadcasting)
tf.subtract(x, y, name=None)   # 减法
tf.multiply(x, y, name=None)   # 乘法
tf.divide(x, y, name=None)     # 浮点除法, 返回浮点数(python3 除法)
tf.mod(x, y, name=None)        # 取余

# 幂指对数操作符:^ ^2 ^0.5 e^ ln 
tf.pow(x, y, name=None)        # 幂次方
tf.square(x, name=None)        # 平方
tf.sqrt(x, name=None)          # 开根号,必须传入浮点数或复数
tf.exp(x, name=None)           # 计算 e 的次方
tf.log(x, name=None)           # 以 e 为底,必须传入浮点数或复数

# 取符号、负、倒数、绝对值、近似、两数中较大/小的
tf.negative(x, name=None)      # 取负(y = -x).
tf.sign(x, name=None)          # 返回 x 的符号
tf.reciprocal(x, name=None)    # 取倒数
tf.abs(x, name=None)           # 求绝对值
tf.round(x, name=None)         # 四舍五入
tf.ceil(x, name=None)          # 向上取整
tf.floor(x, name=None)         # 向下取整
tf.rint(x, name=None)          # 取最接近的整数 
tf.maximum(x, y, name=None)    # 返回两tensor中的最大值 (x > y ? x : y)
tf.minimum(x, y, name=None)    # 返回两tensor中的最小值 (x < y ? x : y)

# 三角函数和反三角函数
tf.cos(x, name=None)    
tf.sin(x, name=None)    
tf.tan(x, name=None)    
tf.acos(x, name=None)
tf.asin(x, name=None)
tf.atan(x, name=None)   

# 其它
tf.div(x, y, name=None)  # python 2.7 除法, x/y-->int or x/float(y)-->float
tf.truediv(x, y, name=None) # python 3 除法, x/y-->float
tf.floordiv(x, y, name=None)  # python 3 除法, x//y-->int
tf.realdiv(x, y, name=None)
tf.truncatediv(x, y, name=None)
tf.floor_div(x, y, name=None)
tf.truncatemod(x, y, name=None)
tf.floormod(x, y, name=None)
tf.cross(x, y, name=None)
tf.add_n(inputs, name=None)  # inputs: A list of Tensor objects, each with same shape and type
tf.squared_difference(x, y, name=None) 

矩阵函数

# 矩阵乘法(tensors of rank >= 2)
tf.matmul(a, b, transpose_a=False, transpose_b=False,    adjoint_a=False, adjoint_b=False, a_is_sparse=False, b_is_sparse=False, name=None)

# 转置,可以通过指定 perm=[1, 0] 来进行轴变换
tf.transpose(a, perm=None, name='transpose')

# 在张量 a 的最后两个维度上进行转置
tf.matrix_transpose(a, name='matrix_transpose')
# Matrix with two batch dimensions, x.shape is [1, 2, 3, 4]
# tf.matrix_transpose(x) is shape [1, 2, 4, 3]

# 求矩阵的迹
tf.trace(x, name=None)

# 计算方阵行列式的值
tf.matrix_determinant(input, name=None)

# 求解可逆方阵的逆,input 必须为浮点型或复数
tf.matrix_inverse(input, adjoint=None, name=None)

# 奇异值分解
tf.svd(tensor, full_matrices=False, compute_uv=True, name=None)

# QR 分解
tf.qr(input, full_matrices=None, name=None)

# 求张量的范数(默认2)
tf.norm(tensor, ord='euclidean', axis=None, keep_dims=False, name=None)

# 构建一个单位矩阵, 或者 batch 个矩阵,batch_shape 以 list 的形式传入
tf.eye(num_rows, num_columns=None, batch_shape=None, dtype=tf.float32, name=None)
# Construct one identity matrix.
tf.eye(2)
==> [[1., 0.],
     [0., 1.]]

# Construct a batch of 3 identity matricies, each 2 x 2.
# batch_identity[i, :, :] is a 2 x 2 identity matrix, i = 0, 1, 2.
batch_identity = tf.eye(2, batch_shape=[3])

# Construct one 2 x 3 "identity" matrix
tf.eye(2, num_columns=3)
==> [[ 1.,  0.,  0.],
     [ 0.,  1.,  0.]]

# 构建一个对角矩阵,rank = 2*rank(diagonal)
tf.diag(diagonal, name=None)
# 'diagonal' is [1, 2, 3, 4]
tf.diag(diagonal) ==> [[1, 0, 0, 0]
                       [0, 2, 0, 0]
                       [0, 0, 3, 0]
                       [0, 0, 0, 4]]
# 其它
tf.diag_part
tf.matrix_diag
tf.matrix_diag_part
tf.matrix_band_part
tf.matrix_set_diag
tf.cholesky
tf.cholesky_solve
tf.matrix_solve
tf.matrix_triangular_solve
tf.matrix_solve_ls
tf.self_adjoint_eig
tf.self_adjoint_eigvals

归约Reduction操作

# 计算输入 tensor 所有元素的和,或者计算指定的轴所有元素的和
tf.reduce_sum(input_tensor, axis=None, keep_dims=False, name=None)
# 'x' is [[1, 1, 1]
#         [1, 1, 1]]
tf.reduce_sum(x) ==> 6
tf.reduce_sum(x, 0) ==> [2, 2, 2]
tf.reduce_sum(x, 1) ==> [3, 3]
tf.reduce_sum(x, 1, keep_dims=True) ==> [[3], [3]]  # 维度不缩减
tf.reduce_sum(x, [0, 1]) ==> 6

# 计算输入 tensor 所有元素的均值/最大值/最小值/积/逻辑与/或
# 或者计算指定的轴所有元素的均值/最大值/最小值/积/逻辑与/或(just like reduce_sum)
tf.reduce_mean(input_tensor, axis=None, keep_dims=False, name=None)
tf.reduce_max(input_tensor, axis=None, keep_dims=False, name=None)
tf.reduce_min(input_tensor, axis=None, keep_dims=False, name=None)
tf.reduce_prod(input_tensor, axis=None, keep_dims=False, name=None)
tf.reduce_all(input_tensor, axis=None, keep_dims=False, name=None)  # 全部满足条件
tf.reduce_any(input_tensor, axis=None, keep_dims=False, name=None) #至少有一个满足条件

-------------------------------------------
# 分界线以上和 Numpy 中相应的用法完全一致
-------------------------------------------

# inputs 为一 list, 计算 list 中所有元素的累计和,
# tf.add(x, y, name=None)只能计算两个元素的和,此函数相当于扩展了其功能
tf.accumulate_n(inputs, shape=None, tensor_dtype=None, name=None)

# Computes log(sum(exp(elements across dimensions of a tensor)))
tf.reduce_logsumexp(input_tensor, axis=None, keep_dims=False, name=None)

# Computes number of nonzero elements across dimensions of a tensor
tf.count_nonzero(input_tensor, axis=None, keep_dims=False, name=None)

Scan

# Compute the cumulative sum of the tensor x along axis
tf.cumsum(x, axis=0, exclusive=False, reverse=False, name=None)
# Eg:
tf.cumsum([a, b, c])  # => [a, a + b, a + b + c]
tf.cumsum([a, b, c], exclusive=True)  # => [0, a, a + b]
tf.cumsum([a, b, c], reverse=True)  # => [a + b + c, b + c, c]
tf.cumsum([a, b, c], exclusive=True, reverse=True)  # => [b + c, c, 0]

# Compute the cumulative product of the tensor x along axis
tf.cumprod(x, axis=0, exclusive=False, reverse=False, name=None)

Segmentation

# Computes the sum/mean/max/min/prod along segments of a tensor
tf.segment_sum(data, segment_ids, name=None)
# Eg:
m = tf.constant([5,1,7,2,3,4,1,3])
s_id = [0,0,0,1,2,2,3,3]
s.run(tf.segment_sum(m, segment_ids=s_id))
>array([13,  2,  7,  4], dtype=int32)

tf.segment_mean(data, segment_ids, name=None)
tf.segment_max(data, segment_ids, name=None)
tf.segment_min(data, segment_ids, name=None)
tf.segment_prod(data, segment_ids, name=None)

# 其它
tf.unsorted_segment_sum
tf.sparse_segment_sum
tf.sparse_segment_mean
tf.sparse_segment_sqrt_n

分割

tf.split(value, num_or_size_splits, axis=0, num=None, name='split')
# 'value' is a tensor with shape [5, 30]
# Split 'value' into 3 tensors with sizes [4, 15, 11] along dimension 1
split0, split1, split2 = tf.split(value, [4, 15, 11], 1)
tf.shape(split0)  # [5, 4]
tf.shape(split1)  # [5, 15]
tf.shape(split2)  # [5, 11]
# Split 'value' into 3 tensors along dimension 1
split0, split1, split2 = tf.split(value, num_or_size_splits=3, axis=1)
tf.shape(split0)  # [5, 10]

tf.slice(input_, begin, size, name=None)
t = tf.constant([[[1, 1, 1], [2, 2, 2]],
                 [[3, 3, 3], [4, 4, 4]],
                 [[5, 5, 5], [6, 6, 6]]])
tf.slice(t, [1, 0, 0], [1, 1, 3])  # [[[3, 3, 3]]]
tf.slice(t, [1, 0, 0], [1, 2, 3])  # [[[3, 3, 3],
                                   #   [4, 4, 4]]]
tf.slice(t, [1, 0, 0], [2, 1, 3])  # [[[3, 3, 3]],
                                   #  [[5, 5, 5]]]

序列比较与索引提取

# 比较两个 list 或者 string 的不同,并返回不同的值和索引
tf.setdiff1d(x, y, index_dtype=tf.int32, name=None)

# 返回 x 中的唯一值所组成的tensor 和原 tensor 中元素在现 tensor 中的索引
tf.unique(x, out_idx=None, name=None)

# x if condition else y, condition 为 bool 类型的,可用tf.equal()等来表示
# x 和 y 的形状和数据类型必须一致
tf.where(condition, x=None, y=None, name=None)

# 返回沿着坐标轴方向的最大/最小值的索引
tf.argmax(input, axis=None, name=None, output_type=tf.int64)
tf.argmin(input, axis=None, name=None, output_type=tf.int64)

# x 的值当作 y 的索引,range(len(x)) 索引当作 y 的值
# y[x[i]] = i for i in [0, 1, ..., len(x) - 1]
tf.invert_permutation(x, name=None)

# 其它
tf.edit_distance

常用函数

tf.concat

把一组向量从某一维上拼接起来,很向numpy中的Concatenate,官网例子:

t1 = [[1, 2, 3], [4, 5, 6]]
t2 = [[7, 8, 9], [10, 11, 12]]
tf.concat([t1, t2], 0) ==> [[1, 2, 3], [4, 5, 6], [7, 8, 9], [10, 11, 12]]
tf.concat([t1, t2], 1) ==> [[1, 2, 3, 7, 8, 9], [4, 5, 6, 10, 11, 12]]
 
# tensor t3 with shape [2, 3]
# tensor t4 with shape [2, 3]
tf.shape(tf.concat([t3, t4], 0)) ==> [4, 3]

如果是list类型的话也是可以的,只要是形似Tensor,最后tf.concat返回的还是Tensor类型

tf.gather

类似于数组的索引,可以把向量中某些索引值提取出来。只适合在一维的情况下使用。

import tensorflow as tf 
 
a = tf.Variable([[1,2,3,4,5], [6,7,8,9,10], [11,12,13,14,15]])
index_a = tf.Variable([0,2])
 
b = tf.Variable([1,2,3,4,5,6,7,8,9,10])
index_b = tf.Variable([2,4,6,8])
 
with tf.Session() as sess:
    sess.run(tf.global_variables_initializer())
    print(sess.run(tf.gather(a, index_a)))
    print(sess.run(tf.gather(b, index_b)))
#  [[ 1  2  3  4  5]
#   [11 12 13 14 15]]

#  [3 5 7 9]

tf.gather_nd

同上,但允许在多维上进行索引。

tf.greater

判断函数。首先张量x和张量y的尺寸要相同,输出的tf.greater(x, y)也是一个和x,y尺寸相同的张量。如果x的某个元素比y中对应位置的元素大,则tf.greater(x, y)对应位置返回True,否则返回False。与此类似的函数还有tf.less、tf.greater_equal。

tf.cast

a = tf.constant([0, 2, 0, 4, 2, 2], dtype='int32')
print(a)
# <tf.Tensor 'Const_1:0' shape=(6,) dtype=int32>
 
b = tf.cast(a, 'float32')
print(b)
# <tf.Tensor 'Cast:0' shape=(6,) dtype=float32>

tf.expand_dims & tf.squeeze

增加 / 压缩张量的维度。

a = tf.constant([0, 2, 0, 4, 2, 2], dtype='int32')
print(a)
# <tf.Tensor 'Const_1:0' shape=(6,) dtype=int32>
 
b = tf.expand_dims(a, 0)
print(b)
# <tf.Tensor 'ExpandDims:0' shape=(1, 6) dtype=int32>
 
print(tf.squeeze(b, 0))
# <tf.Tensor 'Squeeze:0' shape=(6,) dtype=int32>

Tensor的随机存取和遍历

只要能事先知道tensor的size,都可以通过python的循环来对tensor的entry遍历处理。

import tensorflow as tf
 
data=tf.constant([[1,2,3],[4,5,6]])
aa=data*1
size=aa.get_shape()
sum=tf.convert_to_tensor(0)
for i in range(2):
    for j in range(2):
        sum=sum+data[i][j]
with tf.Session() as sess:
    print(sess.run([sum, size]))

但在tensor size只有在运行的时候才能确定时,比如输入不同尺寸的图片,不同数量的bounding box,就没把发在定义graph的时候就确定个数,这时只有使用tf.while_loop。

参考

学习
TensorFlow API

posted @ 2019-01-23 10:55  侯凯  阅读(870)  评论(0编辑  收藏  举报