tensor 维度 问题。

tf.argmax takes two arguments: input and dimension. example: tf.argmx(arr, dimension = 1). or tf.argmax(arr, 1). let arr is ndarray

wrong: Since the indices of array arr are arr[rows, columns], I would expect tf.argmax(arr, 0) to return the index of the maximum element per row, while I would have expected tf.argmax(arr, 1) to return the maximum element per column. Likewise for tf.argmin.

right understand:

Think of the dimension argument of tf.argmax as the axis across which you reduce. 

tf.argmax(arr, 0) reduces across dimension 0, i.e. the rows. Reducing across rows means that you will get the argmax of each individual column.

把tf.argmax变量中的dimension理解为你需要通过轴x来降解维度。 tf.argmax(arr, 0)意味着： 穿过axis=0 （即横轴row）， 穿过row（行）来降解。 而穿过行来讲解，也就是你要对每一列来进行操作。 

This might be counterintuitive, but it falls in line with the conventions used in tf.reduce_max and so on.

Additionally: how does this behave for n-dimensional Tensors? I'm a bit lost at figuring out which dimension relates to reducing i, j, k, l or m in a 5D-Tensor. – daniel451 Jun 29 '16 at 9:12

By definition, if you search for the maximum across rows, you are searching within columns.

For any d-dimensional array, taking the argmax across the ith axis means that, for any possible combination of the d-1 remaining indices, you are searching for the maximum amongst arr[ind1, ind2, ..., ind_i_minus_1, : , ind_i_plus_1, ..., ind_d]  ==》还是对列进行操作， zhi

 下例中 X 定义了有784维的tf 占位符。行数没有定义 。

           Y 定义了有10维的tf 占位符。行数也没有定义 。

X = tf.placeholder("float", [None, 784]) # create symbolic variables
Y = tf.placeholder("float", [None, 10])

w = init_weights([784, 10]) # like in linear regression, we need a shared variable weight matrix for logistic regression

py_x = model(X, w)。# 可以看出py_x 同Y一样是一个有10维的tf占位符。 也就是说有10个标签，10列构成的一个行向量。  行数根据可feed情况推断。

cost = tf.reduce_mean(tf.nn.softmax_cross_entropy_with_logits(py_x, Y)) # compute mean cross entropy (softmax is applied internally)
train_op = tf.train.GradientDescentOptimizer(0.05).minimize(cost) # construct optimizer
predict_op = tf.argmax(py_x, 1) # at predict time, evaluate the argmax of the logistic regression # dimension =1，就是穿过列，对整行进行取最大值对应的行好。 

如果不理解，往下看： 

 

import tensorflow as tf
mygraph = tf.Graph()
with tf.Session() as sess:
x = tf.constant([1,2,3])
y = tf.constant([3,4,5])

op = tf.add(x,y)
result = sess.run(fetches = op)
print result

x = tf.constant([[1,220,55],[4,3,-1]])
with tf.Session() as sess:
result = tf.argmax(x,1) # 输出变量的索引
print sess.run(result) # output: [1 0]
x = tf.constant([[1, 220, 55], [4, 3, -1]])
x_max = tf.reduce_max(x, reduction_indices=[1])
print sess.run(x_max) # ==> "array([220, 4], dtype=int32)"