当我打开一个包时,比如说ggplot2,diamonds我的环境中定义了数据集.
有没有办法访问钻石数据集而不改变我的环境?喜欢
a <- ggplot2.data("diamonds")
Run Code Online (Sandbox Code Playgroud) 这两个元素所在的用户控件具有名为ColumnTypes的属性.
每一元素用相同的表达到主的datacontext相对参考,但第一个广告不工作,而后者则.
你有任何想法如何调查吗?
<DataGrid x:Name="DataGrid" AutoGenerateColumns="False" ItemsSource="{Binding Table}" >
<DataGrid.Columns>
<DataGridComboBoxColumn Header="Type" >
<DataGridComboBoxColumn.ItemsSource>
<Binding Path="DataContext.GetColumnTypes" RelativeSource="{RelativeSource FindAncestor, AncestorType={x:Type UserControl}}" />
</DataGridComboBoxColumn.ItemsSource>
</DataGridComboBoxColumn>
</DataGrid.Columns>
</DataGrid>
<ComboBox Grid.Row="1">
<ComboBox.ItemsSource>
<Binding Path="DataContext.GetColumnTypes" RelativeSource="{RelativeSource FindAncestor, AncestorType={x:Type UserControl}}" />
</ComboBox.ItemsSource>
</ComboBox>
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System.Windows.Data错误:4:无法找到绑定源,引用'RelativeSource FindAncestor,AncestorType ='System.Windows.Controls.UserControl',AncestorLevel ='1''.BindingExpression:路径= DataContext.GetColumnTypes; 的DataItem = NULL; target元素是'DataGridComboBoxColumn'(HashCode = 53813616); target属性是'ItemsSource'(输入'IEnumerable')
我不确定为什么这个简单的代码会导致错误:
object Main {
def main(args: Array[String]) {
val userInterrupted: Future[String] = async {
var inp = await { Future.userInput("")}
"You entered... " + inp
}
}
}
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错误消息:
[error] /Users/reactive programming coursera/nodescala/src/main/scala/nodescala/Main.scala:18: macro has not been expanded
[error] val userInterrupted: Future[String] = async {
[error] ^
[error] one error found
[error] (assignment/compile:compile) Compilation failed
Run Code Online (Sandbox Code Playgroud) 有什么理由说KO不起作用?
type IBase =
abstract member test : (unit * unit) -> unit
type OK() =
interface IBase with
member x.test ((titi,tata)) = () //OK
type KO() =
interface IBase with
member x.test (titi,tata) = () //fail : This override takes a different number of arguments to the corresponding abstract member
Run Code Online (Sandbox Code Playgroud) 在OCaml中,我怎样才能:
在haskell它就像
f arg@{..} = some code using both arg and its fields
Run Code Online (Sandbox Code Playgroud) 我想知道为什么一个例子失败了,而另一个例子却失败了。
(* this fails *)
(* (l fails to type check)
This expression has type 'a but an expression was expected of type
(module M.TFixU)
The module type M.TFixU would escape its scope
*)
let foldList1 (type ar) algr l =
let module M = FixT (ListIntF) in
let (module LU : M.TFixU) = l in
assert false
(* but this works *)
let foldList2 (type ar) algr l =
let (module LU : FixT(ListIntF).TFixU) = l in
assert false …Run Code Online (Sandbox Code Playgroud) 如果我查看M1 计算机的haskell.nix flake 提供的输出,它会开始构建 ghc-8.8.4 等。
\n\xe2\x9d\xaf nix flake show github:input-output-hk/haskell.nix\ngithub:input-output-hk/haskell.nix/1b54ea01568299a0eda578ae9395e20d5c699ee1\n\xe2\x94\x9c\xe2\x94\x80\xe2\x94\x80\xe2\x94\x80checks\n\xe2\x94\x82 \xe2\x94\x9c\xe2\x94\x80\xe2\x94\x80\xe2\x94\x80aarch64-darwin\ntrace: haskell-nix.haskellLib.cleanGit: /nix/store/jmx2m0ldgrjq7p3gb4yyca47nvbvspfl-source does not seem to be a git repository,\nassuming it is a clean checkout.\ntrace: No index state specified for haskell-project, using the latest index state that we know about (2022-02-07T00:00:00Z)!\ntrace: No index state specified for haskell-project, using the latest index state that we know about (2022-02-07T00:00:00Z)!\ntrace: No index state specified for haskell-project, using the latest index state that we know about (2022-02-07T00:00:00Z)!\ntrace: WARNING: No materialized …Run Code Online (Sandbox Code Playgroud) 文档flake-utils有以下示例作为文档
eachSystem [ system.x86_64-linux ] (system: { hello = 42; })
# => { hello = { x86_64-linux = 42; }; }
eachSystem allSystems (system: { hello = 42; })
# => {
hello.aarch64-darwin = 42,
hello.aarch64-genode = 42,
hello.aarch64-linux = 42,
...
hello.x86_64-redox = 42,
hello.x86_64-solaris = 42,
hello.x86_64-windows = 42
}
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据我所知,必须
> nix repl
nix-repl> e = builtins.getFlake("github:numtide/flake-utils")
nix-repl> with e.outputs.lib;
eachSystem [ system.x86_64-linux ] (system: { hello = 42; })
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获取结果值(也可以执行:a e.outputs.lib“将结果集中的属性添加到范围”而不使用该 …
我是Python的新手,无法理解为什么这样的东西不起作用.我也找不到其他地方提出的问题.
toto = {'a':1, 'c':2 , 'b':3}
toto.keys().sort() #does not work (yields none)
(toto.keys()).sort() #does not work (yields none)
eval('toto.keys()').sort() #does not work (yields none)
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然而,如果我检查类型,我看到我在列表上调用sort(),那么问题是什么..
toto.keys().__class__ # yields <type 'list'>
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我有这个工作的唯一方法是添加一些丑陋的临时变量
temp = toto.keys()
temp.sort()
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我在这里缺少什么,必须有一个更好的方法来做到这一点.