小编Abh*_*yal的帖子

如果输入为零,我想创建一个如下所示的数组:

[1,0,0,0,0,0,0,0,0,0]

如果输入为5:

[0,0,0,0,0,1,0,0,0,0]

对于上面我写道:

np.put(np.zeros(10),5,1)

但它不起作用.

有什么方法可以在一行中实现吗？

我从源(文档)安装tensorflow .

Cuda驱动版:

nvcc: NVIDIA (R) Cuda compiler driver
Cuda compilation tools, release 7.5, V7.5.17

当我运行以下命令时:

bazel-bin/tensorflow/cc/tutorials_example_trainer --use_gpu

它给了我以下错误:

I tensorflow/stream_executor/dso_loader.cc:108] successfully opened CUDA library libcublas.so locally
I tensorflow/stream_executor/dso_loader.cc:108] successfully opened CUDA library libcudnn.so locally
I tensorflow/stream_executor/dso_loader.cc:108] successfully opened CUDA library libcufft.so locally
I tensorflow/stream_executor/dso_loader.cc:108] successfully opened CUDA library libcuda.so.1 locally
I tensorflow/stream_executor/dso_loader.cc:108] successfully opened CUDA library libcurand.so locally
I tensorflow/stream_executor/cuda/cuda_gpu_executor.cc:925] successful NUMA node read from SysFS had negative value (-1), but there must be at least one …

这是我正在运行的示例MNIST代码:

from tensorflow.examples.tutorials.mnist import input_data
mnist = input_data.read_data_sets('MNIST_data', one_hot=True)

import tensorflow as tf
sess = tf.InteractiveSession()

x = tf.placeholder(tf.float32, shape=[None, 784])
y_ = tf.placeholder(tf.float32, shape=[None, 10])

W = tf.Variable(tf.zeros([784,10]))
b = tf.Variable(tf.zeros([10]))

y = tf.nn.softmax(tf.matmul(x,W) + b)

def weight_variable(shape):
  initial = tf.truncated_normal(shape, stddev=0.1)
  return tf.Variable(initial)

def bias_variable(shape):
  initial = tf.constant(0.1, shape=shape)
  return tf.Variable(initial)


def conv2d(x, W):
  return tf.nn.conv2d(x, W, strides=[1, 1, 1, 1], padding='SAME')

def max_pool_2x2(x):
  return tf.nn.max_pool(x, ksize=[1, 2, 2, 1],
                        strides=[1, 2, 2, 1], padding='SAME')


W_conv1 = weight_variable([5, …

我试图弄清楚为什么这不起作用(游乐场):

fn main() {
    let a = vec![1, 2, 3, 4];
    let b = a.clone();
    // slice and iter (wrong way)
    let s: i32 = &a[1..a.len()].iter()
        .zip(&b[1..b.len()].iter())
        .map(|(x, y)| x * y)
        .sum();
    println!("{}", s);
}

错误:

rustc 1.13.0 (2c6933acc 2016-11-07)
error[E0277]: the trait bound `&std::slice::Iter<'_, {integer}>: std::iter::Iterator` is not satisfied
 --> <anon>:6:10
  |
6 |         .zip(&b[1..b.len()].iter())
  |          ^^^ trait `&std::slice::Iter<'_, {integer}>: std::iter::Iterator` not satisfied
  |
  = note: `&std::slice::Iter<'_, {integer}>` is not an iterator; maybe try calling `.iter()` …

我正在尝试按列获取唯一计数，但我的数组具有分类变量（dtype 对象）

val, count = np.unique(x, axis=1, return_counts=True)

虽然我收到这样的错误：

TypeError: The axis argument to unique is not supported for dtype object

我该如何解决这个问题？

样品 x：

array([[' Private', ' HS-grad', ' Divorced'],
       [' Private', ' 11th', ' Married-civ-spouse'],
       [' Private', ' Bachelors', ' Married-civ-spouse'],
       [' Private', ' Masters', ' Married-civ-spouse'],
       [' Private', ' 9th', ' Married-spouse-absent'],
       [' Self-emp-not-inc', ' HS-grad', ' Married-civ-spouse'],
       [' Private', ' Masters', ' Never-married'],
       [' Private', ' Bachelors', ' Married-civ-spouse'],
       [' Private', ' Some-college', ' Married-civ-spouse']], dtype=object)

需要以下计数： …

我试图在张量流上写一个分布式变分自动编码器standalone mode.

我的群集包括3台机器,命名为m1,m2和m3.我试图在m1上运行1 ps服务器,在m2和m3上运行2个工作服务器.(分布式tensorflow文档中的示例培训师程序)在m3上,我收到以下错误消息:

Traceback (most recent call last): 
 File "/home/yama/mfs/ZhuSuan/examples/vae.py", line 241, in <module> 
   save_model_secs=600) 
 File "/mfs/yama/tensorflow/local/lib/python2.7/site-packages/tensorflow/python/training/supervisor.py", line 334, in __init__ 
   self._verify_setup() 
 File "/mfs/yama/tensorflow/local/lib/python2.7/site-packages/tensorflow/python/training/supervisor.py", line 863, in _verify_setup 
   "their device set: %s" % op) 
ValueError: When using replicas, all Variables must have their device set: name: "Variable"
op: "Variable" 
attr { 
 key: "container" 
 value { 
   s: "" 
 } 
} 
attr { 
 key: "dtype" 
 value { 
   type: DT_INT32 
 } 
} 
attr { 
 key: "shape" 
 value { …

说我有以下三个列表：

aList = [1,2,3,4,5,6]
bList = ['a','b','c','d']
cList = [1,2]

我想使用来遍历它们zip。

通过如下方式使用循环zip：

from itertools import cycle
for a,b,c in zip(aList, cycle(bList), cycle(cList)):
    print a,b,c

我得到的结果是：

1 a 1
2 b 2
3 c 1
4 d 2
5 a 1
6 b 2

虽然我希望我的结果像：

1 a 1
2 b 1
3 c 1
4 d 1
5 a 2
6 b 2

我正在尝试切片矢量并在Rust中同时打印它.这是我的代码:

fn main() {
    let a = vec![1, 2, 3, 4];
    println!("{:?}", a[1..2]);
}

错误:

error[E0277]: the trait bound `[{integer}]: std::marker::Sized` is not satisfied
 --> src/main.rs:6:5
  |
6 |     println!("{:?}", a[1..3]);
  |     ^^^^^^^^^^^^^^^^^^^^^^^^^^ trait `[{integer}]: std::marker::Sized` not satisfied
  |
  = note: `[{integer}]` does not have a constant size known at compile-time
  = note: required by `std::fmt::ArgumentV1::new`
  = note: this error originates in a macro outside of the current crate

如何打印此切片矢量？

我正在绘制不同地图上设备的每小时更新。为了动画和分析更新，我通过清除 jupyter 中单元格的输出在每次迭代中显示一个新图。

map_ = [[] for i in range(max(dfByDeviceidHourly['hour'])+1)]
for hour in dfByDeviceidHourly['hour'].unique():
    df_map = dfByDeviceidHourly[dfByDeviceidHourly['hour'] == hour]
    map_[hour] = folium.Map(location=list(df_map[['geocoordinates_latitude','geocoordinates_longitude']].mean()), zoom_start=13)
    for i,mark in enumerate(df_map):
        folium.Marker(list(df_map[['geocoordinates_latitude','geocoordinates_longitude']].iloc[i]), popup='The Waterfront').add_to( map_[hour] )
    clear_output(wait = True)
    display(map_[hour])
    time.sleep(4)

有什么方法可以让我从上面制作视频或 gif 吗？

我找到了一种方法（这里：https : //github.com/python-visualization/folium/issues/35），它涉及将图像保存为 html，将它们转换为 png，从中创建一个 gif，然后在笔记本。尽管我正在为它寻找更强大的替代方案。

我想在我的中心对齐输出(包括文本和图表)ipython notebook.有没有一种方法可以在同一个笔记本中添加样式？代码或屏幕截图示例将有很大帮助.

我正在使用手电筒摘要

from torchsummary import summary

我想在打印模型摘要时传递多个参数，但是这里提到的示例：pytorch中的模型摘要仅采用了一个参数。例如：

model = Network().to(device)
summary(model,(1,28,28))

原因是 forward 函数需要两个参数作为输入，例如：

def forward(self, img1, img2):

我如何在这里传递两个参数？

我想删除剧情边界内的额外空间

plt.boxplot(parkingData_agg['occupancy'], 0, 'rs', 0, 0.75)
plt.tight_layout() # This didn't work. Maybe it's not for the purpose I am thinking it is used for.
plt.yticks([0],['Average Occupancy per slot'])
fig = plt.figure(figsize=(5, 1), dpi=5) #Tried to change the figsize but it didn't work
plt.show()

所需的图如下图左图所示