我正试图用BNFC编写一个编译器。我将使用BNFC来生成抽象语法树。但是,我得到了错误,我似乎不知道为什么。它似乎没有太多的文档。
以下是我所犯的错误:
Bad coercion in rule _. Prog ::= Block
Bad coercion in rule _. Declarations ::= Declaration ";" Declarations
Bad coercion in rule _. Declarations ::=
Bad coercion in rule _. Declaration ::= Var_declaration
Bad coerc
Edward Kmett的实验性提供了各种解除强制的实用程序,我已经在这个问题的末尾粘贴了一些实用程序。包中的关键类是
class Representational (t :: k1 -> k2) where
-- | An argument is representational if you can lift a coercion of the argument into one of the whole
rep :: Coercion a b -> Coercion (t a) (t b)
给定类型
newtype Fix p a = In {out :: p (Fi
我有一个数据框架,其中包含“日期变量”。(测试数据和代码可用)
但是,我使用"function = caretFunc“。它显示错误信息。
Error in { : task 1 failed - "missing value where TRUE/FALSE needed"
In addition: Warning messages:
1: In FUN(newX[, i], ...) : NAs introduced by coercion
2: In FUN(newX[, i], ...) : NAs introduced by coercion
3: I
I ^有一个如下所示的数据集:
typestudy dloop cytb coi other microsat SNP
methods no no no no yes no
methods yes no no no no yes
methods no no no no yes no
methods no no no no yes no
wildcrime no no no yes no no
taxonomy no no no no yes
我试图运行以下代码:
marka <-c("Skoda","Skoda","Opel","Volkswagen","Toyota","Toyota","Ford","Dacia","Skoda","Volkswagen","Nissan","Renault","Hyundai","Fiat","Skoda","Toyota","
带有关联数据族X的类型类X需要一个函数coerceX。如果我像下面这样实现了类型分类,我如何编写coerceX
{-# LANGUAGE ScopedTypeVariables #-}
{-# LANGUAGE TypeFamilies #-}
{-# LANGUAGE TypeApplications #-}
import Data.Type.Coercion
import Control.Monad.Reader
data (T r t)
class C t where
data X t :: * -> *
coerceX :: Coercion a b -> C
尝试将xlsx文件加载到R时,使用openxlsx::loadWorkbook
ExcelFile <- loadWorkbook(ReportFilePath) #ReportFilePath is a character variable containing path to the file
R返回警告:
In sprintf("<Relationship Id=\"rId4\" Type=\"http://schemas.openxmlformats.org/officeDocument/2006/relationships/external
我遇到了一种情况(在记录各种数据更改时),我需要确定引用是否具有有效的字符串强制(例如,可以正确地打印到日志中或存储在数据库中)。在中没有任何东西可以做到这一点,所以我使用该库中的其他方法拼凑了一些东西:
use strict;
use warnings;
use Scalar::Util qw(reftype refaddr);
sub has_string_coercion
{
my $value = shift;
my $as_string = "$value";
my $ref = ref $value;
my $reftype =
我花了很多时间寻找它,但是找不到。如果这是一个基本问题,请不要对我大加抨击:)
我想用下面的向量生成一个散点图
> x
[1] "a" "b" "c" "d"
> y
[1] 5 6 3 4
我使用了xyplot,但它给出了以下错误
> xyplot(y~x)
Hit <Return> to see next plot:
Warning messages:
1: In order(as.numeric(x)) : NAs introduced by coercion
2: In diff(as.num
我试图从表文件中提取一个列的范围,使用列名来指定范围。我试过了
dayTable[c("Peter":"Michael")]
并得到了
Error in "Peter":"Michael" : NA/NaN argument
In addition: Warning messages:
1: In `[.data.frame`(dayTable, c("Peter":"Michael")) :
NAs introduced by coercion
2: In `[.data.frame`(day
如何获取变量Ticker等于"AA“的所有行?数据结构是tibble。我的数据存储在返点数据中,如下所示
A tibble: 1,048,575 x 7
Date Ticker CUR CON REBATERATE FEERATE AVAILABLE
<chr> <chr> <chr> <dbl> <dbl> <dbl> <dbl>
1 09/25/2017 A USD 1715006 0.
当我尝试创建下面的简单图形时,为什么会出现错误?如果我用数字替换"a“和"b”,那么它可以工作吗?任何解决方案
g1 <- graph(c("a","b"),directed=TRUE)
错误是
Error in graph(c("a", "b"), directed = TRUE) :
At type_indexededgelist.c:117 : cannot create empty graph with negative number of vertices, Invalid value
In
IAutomation -如何检索通过指针传递的参数值?
在写入的.idl文件中:
interface IAutomation : IDispatch {
[id(0x000000e0), helpstring("Returns if the player is ready. Many of the other commands fail as long as the player is not ready.")]
HRESULT GetReady([out] VARIANT* Ready);
};
我想应该是GetReady()方法,而不是属性
假设我想循环一个列表到列表的第10个元素。你能告诉我为什么代码不能工作吗? loop <- function (list) {
for (i in (list:list[[10]])) {
df_i <- import(paste0(i))
}
}
loop(list) 它返回: Error in ord_dirs:ord_dirs[[10]] : NA/NaN argument
In addition: Warning messages:
1: In ord_dirs:ord_dirs[[10]] :
numerical expression has 18
你能修复这个错误吗:
Parameter Arg: Type.
Parameter F X XP: Arg.
Parameter Sen Phy Leg Inf: Arg -> Prop.
Parameter tree car: Phy X.
Parameter mary john: Phy XP /\ Leg XP /\ Sen XP.
Fail Coercion c (u:Arg) (x y z: Arg -> Prop) (t:x u /\ y u /\ z u): x u := fun t => @proj1 (x u) (y u /\ z u) t.
(*The t