coercible | Powerful , flexible and configurable coercion library
kandi X-RAY | coercible Summary
kandi X-RAY | coercible Summary
Powerful, flexible and configurable coercion library. And nothing more.
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Top functions reviewed by kandi - BETA
- Determine the class of the given type
- Defines an option
- Initialize a new instance
- Determine the primitive
- Defines all options for the given options .
- Returns a hash with the default options .
- Determine the given string
- Add valid options .
- Extracts the options from the given class .
- Sets the options for the given hash .
coercible Key Features
coercible Examples and Code Snippets
Community Discussions
Trending Discussions on coercible
QUESTION
I have this data
...ANSWER
Answered 2021-Jun-04 at 17:43We need to specify the column name as unquoted
QUESTION
I am working with R. Using a tutorial, I was able to create a statistical model and produce visual plots for some of the outputs:
...ANSWER
Answered 2021-May-20 at 01:23As per the ggplot2 documentation, you need to provide a data.frame()
or object that can be converted (coerced) to a data.frame()
. In this case, if you want to reproduce the plot above in ggplot2, you will need to manually set up the data frame yourself.
Below is an example of how you could set up the data to display the plot in ggplot2.
Data FrameFirst we create a data.frame()
with the variables that we want to plot. The easiest way to do this is to just group them all in as separate columns. Note that I have used the as.numeric()
function to first coerce the predicted values to a vector, because they were previously a data.table
row, and if you don't convert them they are maintained as rows.
QUESTION
I want to associate _
with coerce
yet I can't give a type signature to it.
Is there any trickery to get around that ?
...ANSWER
Answered 2021-May-17 at 15:54_
is a reserved name which can not be redefined. It can be used in patterns as a wildcard, e.g.
QUESTION
Though I tried to search whether it is duplicate, but I cannot find similar question. (though a similar one is there, but that is somewhat different from my requirement)
My question is that whether we can use string manipulation function such substr
or stringr::str_remove
inside .names
argument of dplyr::across
. As a reproducible example consider this
ANSWER
Answered 2021-May-15 at 15:46This works, but with probably a few caveats. You can use functions inside a glue specification, so you could clean up the strings that way. However, when I tried escaping the "."
, I got an error, which I assume has something to do with how across
parses the string. If you need something more dynamic, you might want to dig into the source code at that point.
In order to use the {.fn}
helper, at least in conjunction with creating the glue string on the fly like this, the function needs a name; otherwise you get a number for the function's index in the .fns
argument. I tested this out with a second function and using lst
for automatic naming.
QUESTION
In the current implementation of Coercible
can we quantify over "representational preserving type constructors", to extract a safer proof in the code below
ANSWER
Answered 2021-May-06 at 17:16We can not allow that without making the type system unsafe.
Assume we have a safe way to obtain
QUESTION
I digitized some roads as multilines, hospitals as multipoints, boundary as polygon, then created how many roads intersect using the Simple Features (SF) library by getting latitudes and longitudes from google maps and plotted it using ggplot2 it worked well.
I then wanted to check and plot how many roads intersects with a hospital and created a 200mtr buffer around it and tried using st_intersects() function for the same, using this only gave 1:1 as answer and a message saying
Sparse geometry binary predicate list of length 1, where the predicate was `intersects' 1: 1
And when I tried plotting it, using ggplot it gives this error message
Error:
data
must be a data frame, or other object coercible byfortify()
, not an S3 object with class sgbp/list Runrlang::last_error()
to see where the error occurred.
I am pasting my code below,
...ANSWER
Answered 2021-May-03 at 19:56In your code you are using st_intersects
to create the road features. But this function returns a booolean, not a spatial feature. Try st_intersection()
instead.
QUESTION
I am trying to create a histogram for the wage variable
My code is
...ANSWER
Answered 2021-Apr-13 at 21:02Try this. Last line: I think it was typo: changed logW with W.
QUESTION
I want to code a ggplot2
visualization as a function, and then apply
the function on each row of a dataframe (I want to use apply
to avoid a for
loop, as suggested here.)
The data:
...ANSWER
Answered 2021-Apr-09 at 17:10As suggested by Gregor Thomas in the comment:
QUESTION
I am trying to graph the residual effects of the mixed effects model using the ggplot2 function. However, after performing a search I found some functions available but what seems to me is that for the function nlme they are not working.
The graphs that I intend to make are those of the example below:
The data is here.
Data: https://drive.google.com/file/d/19mykz4B7jkTilbtwPQb3NUI09YZwohhs/view?usp=sharing
The computational routines that I tried initially are below, see the errors that are appearing when executing the function in ggplot2.
...ANSWER
Answered 2021-Feb-24 at 13:02[SOLUTION]
QUESTION
I am trying to visualize qualitative grade based data, and I would like to order the grades from worst to best. The current ordering seems to be lexicographic one. How can I change this? Unfortunately I do not know how to share a working example: the column G3.x
is full of letter based qualitative grades. When I tried the code with an example test_data <- c("Excellent", "Very Good", "Good", "Sufficient", "Weak", "Poor")
I get errors: Error: data must be a data frame, or other object coercible by fortify(), not a character vector
ANSWER
Answered 2021-Mar-20 at 13:52The ordering of any discrete item is done in the following way:
- if your column is a factor,
ggplot2
orders by the levels of the factor - if your column is not a factor,
ggplot2
orders alphanumerically
In this case, the solution is to set the levels of math_and_portuguese$G3.x
discretely by converting to a factor and specifying the levels you want:
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