In r language, what is the meaning of "str" in "ls.str()"? I understand that ls.str() gives you a detailed description of the objects in the active memory. But I am still confused about what str stands for then?
From ?ls.str:
‘ls.str’ and ‘lsf.str’ are variations of ‘ls’ applying ‘str()’
From ?str:
Compactly display the internal *str*ucture of an R object ...
so the answer is "list structures".
Taken from the Full reference manual.
str
Compactly Display the Structure of an Arbitrary R Object
Description
Compactly display the internal
str
ucture of an
R
object, a diagnostic function and an alternative to
summary
(and to some extent,
dput
). Ideally, only one line for each ‘basic’ structure is displayed. It
is especially well suited to compactly display the (abbreviated) contents of (possibly nested) lists.
The idea is to give reasonable output for
any
R
object. It calls
args
for (non-primitive) function
objects.
strOptions()
is a convenience function for setting
options
(str = .)
, see the examples.
The abbreviation "str" is taken from the first 3 letters of the word structure.
[The same is highlighted in the official R documentation below]
https://www.rdocumentation.org/packages/utils/versions/3.5.1/topics/str
Related
I'm trying to rename a specific column in my R script using the colnames function but with no sucess so far.
I'm kinda new around programming so it may be something simple to solve.
Basically, I'm trying to rename a column called Reviewer Overall Notes and name it Nota Final in a data frame called notas with the codes:
colnames(notas$`Reviewer Overall Notes`) <- `Nota Final`
and it returns to me:
> colnames(notas$`Reviewer Overall Notes`) <- `Nota Final`
Error: object 'Nota Final' not found
I also found in [this post][1] a code that goes:
colnames(notas) [13] <- `Nota Final`
But it also return the same message.
What I'm doing wrong?
Ps:. Sorry for any misspeling, English is not my primary language.
You probably want
colnames(notas)[colnames(notas) == "Reviewer Overall Notes"] <- "Nota Final"
(#Whatif's answer shows how you can do this with the numeric index, but probably better practice to do it this way; working with strings rather than column indices makes your code both easier to read [you can see what you're renaming] and more robust [in case the order of columns changes in the future])
Alternatively,
notas <- notas %>% dplyr::rename(`Nota Final` = `Reviewer Overall Notes`)
Here you do use back-ticks, because tidyverse (of which dplyr is a part) prefers its arguments to be passed as symbols rather than strings.
Why using backtick? Use the normal quotation mark.
colnames(notas)[13] <- 'Nota Final'
This seems to matter:
df <- data.frame(a = 1:4)
colnames(df)[1] <- `b`
Error: object 'b' not found
You should not use single or double quotes in naming:
I have learned that we should not use space in names. If there are spaces in names (it works and is called a non-syntactic name: And according to Wickham Hadley's description in Advanced R book this is due to historical reasons:
"You can also create non-syntactic bindings using single or double quotes (e.g. "_abc" <- 1) instead of backticks, but you shouldn’t, because you’ll have to use a different syntax to retrieve the values. The ability to use strings on the left hand side of the assignment arrow is an historical artefact, used before R supported backticks."
To get an overview what syntactic names are use ?make.names:
make.names("Nota Final")
[1] "Nota.Final"
In Julia, how can I check an English word is a meaningful word? Suppose I want to know whether "Hello" is meaningful or not. In Python, one can use the enchant or nltk packages(Examples: [1],[2]). Is it possible to do this in Julia as well?
What I need is a function like this:
is_english("Hello")
>>>true
is_english("Hlo")
>>>false
# Because it doesn't have meaning! We don't have such a word in English terminology!
is_english("explicit")
>>>true
is_english("eeplicit")
>>>false
Here is what I've tried so far:
I have a dataset that contains frequent 5char English words(link to google drive). So I decided to augment it to my question for better understanding. Although this dataset is not adequate (because it just contains frequent 5char meaningful words, not all the meaningful English words with any length), it's suitable to use it to show what I want:
using CSV
using DataFrames
df = CSV.read("frequent_5_char_words.csv" , DataFrame , skipto=2)
df = [lowercase(item) for item in df[:,"0"]]
function is_english(word::String)::Bool
return lowercase(word) in df
end
Then when I try these:
julia>is_english("Helo")
false
julia>is_english("Hello")
true
But I don't have an affluent dataset! So this isn't enough. So I'm curious if there are any packages like what I mentioned before, in Julia or not?
(not enough rep to post a comment!)
You can still use NLTK in Julia via PyCall. Or, as it seems you don't need an NLP tool but just a dictionary, you can use wiktionary to do some lookup or build the dataset.
There is a recently new package, Named LanguageDetect.jl. It does not return true/false, but a list of probabilities. You could define something like:
using LanguageDetect: detect
function is_english(text, threshold=0.8)
langs = detect(text)
for lang in langs
if lang.language == "en"
return lang.probability >= threshold
end
end
ret
I thought that R did not do partial matching on named lists, so I'm confused by the example below. I tried reading the Argument matching document but I'm still not sure of what's going on. Any help understanding this example would be appreciated.
ll <- list("dir_session" = "some_directory")
print(ll$dir_session) # prints contents of ll$dir_session as expected
print(ll$dir) # prints contents of ll$dir_session, but I expected to print NULL
print(ll[["dir"]]) # prints NULL as expected
Not sure if it makes a difference but I'm using R version 3.3.3 (2017-03-06).
I'm afraid the answer is you thought wrong. It has less to do with the class of object (a named list) and more to do with the "$" operator which does partial matching. See the ?Extract help page. This is different than argument matching when calling a function.
My question refers to redundant code and a problem that I've been having with a lot of my R-Code.
Consider the following:
list_names<-c("putnam","einstein","newton","kant","hume","locke","leibniz")
combined_df_putnam$fu_time<-combined_df_putnam$age*365.25
combined_df_einstein$fu_time<-combined_einstein$age*365.25
combined_df_newton$fu_time<-combined_newton$age*365.25
...
combined_leibniz$fu_time<-combined_leibniz$age*365.25
I am trying to slim-down my code to do something like this:
list_names<-c("putnam","einstein","newton","kant","hume","locke","leibniz")
paste0("combined_df_",list_names[0:7]) <- data.frame("age"=1)
paste0("combined_df_",list_names[0:7]) <- paste0("combined_df_",list_names[0:7])$age*365.25
When I try to do that, I get "target of assignment expands to non-language object".
Basically, I want to create a list that contains descriptors, use that list to create a list of dataframes/lists and use these shortcuts again to do calculations. Right now, I am copy-pasting these assignments and this has led to various mistakes because I failed to replace the "name" from the previous line in some cases.
Any ideas for a solution to my problem would be greatly appreciated!
The central problem is that you are trying to assign a value (or data.frame) to the result of a function.
In paste0("combined_df_",list_names[0:7]) <- data.frame("age"=1), the left-hand-side returns a character vector:
> paste0("combined_df_",list_names[0:7])
[1] "combined_df_putnam" "combined_df_einstein" "combined_df_newton"
[4] "combined_df_kant" "combined_df_hume" "combined_df_locke"
[7] "combined_df_leibniz"
R will not just interpret these strings as variables that should be created and be referenced to. For that, you should look at the function assign.
Similarily, in the code paste0("combined_df_",list_names[0:7])$age*365.25, the paste0 function does not refer to variables, but simply returns a character vector -- for which the $ operator is not accepted.
There are many ways to solve your problem, but I will recommend that you create a function that performs the necessary operations of each data frame. The function should then return the data frame. You can then re-use the function for all 7 philosophers/scientists.
I'm having trouble working with a data table in R. This is probably something really simple but I can't find the solution anywhere.
Here is what I have:
Let's say t is the data table
colNames <- names(t)
for (col in colNames) {
print (t$col)
}
When I do this, it prints NULL. However, if I do it manually, it works fine -- say a column name is "sample". If I type t$"sample" into the R prompt, it works fine. What am I doing wrong here?
You need t[[col]]; t$col does an odd form of evaluation.
edit: incorporating #joran's explanation:
t$col tries to find an element literally named 'col' in list t, not what you happen to have stored as a value in a variable named col.
$ is convenient for interactive use, because it is shorter and one can skip quotation marks (i.e. t$foo vs. t[["foo"]]. It also does partial matching, which is very convenient but can under unusual circumstances be dangerous or confusing: i.e. if a list contains an element foolicious, then t$foo will retrieve it. For this reason it is not generally recommended for programming.
[[ can take either a literal string ("foo") or a string stored in a variable (col), and does not do partial matching. It is generally recommended for programming (although there's no harm in using it interactively).