我正在使用python龟库绘制一个大图.无论何时绘制,显示的区域都是中心(即滚动条位于中间位置).我想滚动到左上角区域.有没有办法做到这一点?
我能够将一个常量乘以一个数组,但无法对除法运算符执行相同的操作。预期的?
julia> 2 * [1,2,3]
3-element Array{Int64,1}:
2
4
6
julia> 2 / [1,2,3]
ERROR: MethodError: no method matching /(::Int64, ::Array{Int64,1})
Closest candidates are:
/(::Union{Int128, Int16, Int32, Int64, Int8, UInt128, UInt16, UInt32, UInt64, UInt8}, ::Union{Int128, Int16, Int32, Int64, Int8, UInt128, UInt16, UInt32, UInt64, UInt8}) at int.jl:38
/(::Union{Int16, Int32, Int64, Int8, UInt16, UInt32, UInt64, UInt8}, ::BigInt) at gmp.jl:381
/(::T<:Integer, ::T<:Integer) where T<:Integer at int.jl:36
...
Run Code Online (Sandbox Code Playgroud) 我刚刚尝试在Google Colab中使用TPU,我想看看TPU比GPU快多少。我惊讶地得到了相反的结果。
以下是NN。
random_image = tf.random_normal((100, 100, 100, 3))
result = tf.layers.conv2d(random_image, 32, 7)
result = tf.reduce_sum(result)
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性能结果:
CPU: 8s
GPU: 0.18s
TPU: 0.50s
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我不知道为什么。...TPU的完整代码如下:
def calc():
random_image = tf.random_normal((100, 100, 100, 3))
result = tf.layers.conv2d(random_image, 32, 7)
result = tf.reduce_sum(result)
return result
tpu_ops = tf.contrib.tpu.batch_parallel(calc, [], num_shards=8)
session = tf.Session(tpu_address)
try:
print('Initializing global variables...')
session.run(tf.global_variables_initializer())
print('Warming up...')
session.run(tf.contrib.tpu.initialize_system())
print('Profiling')
start = time.time()
session.run(tpu_ops)
end = time.time()
elapsed = end - start
print(elapsed)
finally:
session.run(tf.contrib.tpu.shutdown_system())
session.close()
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试图关注http://www.telerik.com/kendo-angular-ui/getting-started/
从浏览器控制台得到此错误...服务器端没有错误...
<button kendoButton (click)="onButtonClick()" [ERROR ->][primary]=true >My Kendo UI Button</button>
"): AppComponent@9:46 ; Zone: <root> ; Task: Promise.then ; Value: Error: Template parse errors:(…) Error: Template parse errors:
Can't bind to 'primary' since it isn't a known property of 'button'. ("
<app-date>test</app-date>
<button kendoButton (click)="onButtonClick()" [ERROR ->][primary]=true >My Kendo UI Button</button>
"): AppComponent@9:46
at TemplateParser.parse (http://localhost:4200/main.bundle.js:15261:19)
at RuntimeCompiler._compileTemplate (http://localhost:4200/main.bundle.js:33578:51)
at http://localhost:4200/main.bundle.js:33501:83
at Set.forEach (native)
at compile (http://localhost:4200/main.bundle.js:33501:47)
at ZoneDelegate.invoke (http://localhost:4200/main.bundle.js:64762:28)
at Zone.run (http://localhost:4200/main.bundle.js:64655:43)
at http://localhost:4200/main.bundle.js:65021:57
at ZoneDelegate.invokeTask (http://localhost:4200/main.bundle.js:64795:37) …Run Code Online (Sandbox Code Playgroud) 我升级了 Linux 内核,但 dovecot 无法启动,并出现以下错误消息:
Error: service(managesieve-login): listen(*, 4190) failed: Address already in use
Error: service(pop3-login): listen(*, 110) failed: Address already in use
Error: service(pop3-login): listen(*, 995) failed: Address already in use
Error: service(imap-login): listen(*, 143) failed: Address already in use
Error: service(imap-login): listen(*, 993) failed: Address already in use
Fatal: Failed to start listeners
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奇怪的是,我找不到任何与这些端口号绑定的进程。以下所有命令均不返回任何内容。
# netstat -tulpn | grep 110
# ss -tulpn |grep 110
# fuser 110/tcp
# lsof -i :110
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我也尝试将设置更改listen为我的特定 IP 地址,但仍然失败。
知道我该如何解决这个问题吗?这是我的版本信息: …
假设我有这两个数据框:
>>> df1 = pd.DataFrame({'grp':[1,1,2], 'x':[6,4,2], 'y':[7,8,9]})
>>> df1
grp x y
0 1 6 7
1 1 4 8
2 2 2 9
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>>> df2 = pd.DataFrame({'grp':[1], 'x':[6], 'z':[3]})
>>> df2
grp x z
0 1 6 3
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我认为半连接可以用单列轻松完成,例如
>>> df1[df1.grp.isin(df2.grp)]
grp x y
0 1 6 7
1 1 4 8
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问题是:如何使用两列 -grp和 来做到这一点x?
以下代码无法编译:
auto greater_than = [](const int a, const int b) { return a > b; };
auto is_adult = bind(&greater_than, placeholders::_1, 17);
cout << "Age 19 is adult = " << is_adult(19) << endl;
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该is_adult呼叫是给一个模糊的错误:
error: no matching function for call to object of type 'std::_Bind<(lambda at main.cpp:62:23) *(std::_Placeholder<1>, int)>'
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但是,如果我将其移动greater_than为全局函数,则它可以工作。
这是为什么?
在C ++中,可以创建通过一个子类public,protected或private继承。在 UML 类图中表示这一点的符号是什么?我正在考虑在箭头上贴上标签,但不确定这是否是常见做法。
c++ ×2
angular ×1
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