I am a beginner and don't understand how to not produce an infinite loop in this context. The code is Python. I do not believe I am expected to produce code that automatically put the names in alphabetical order, rather its just to get comfortable printing characters.
Put all of the names in an array first, then sort, then print.
names = ['Simeoni', 'Juhani',...]
names.sort()
print(names)
Related
I just hope to learn how to make a simple statistical summary of the random numbers fra row 1 to 5 in R. (as shown in picture).
And then assign these rows to a single variable.
enter image description here
Hope you can help!
When you type something like 3 on a single line and ask R to "run" it, it doesn't store that anywhere -- it just evaluates it, meaning that it tries to make sense out of whatever you've typed (such as 3, or 2+1, or sqrt(9), all of which would return the same value) and then it more or less evaporates. You can think of your lines 1 through 5 as behaving like you've used a handheld scientific calculator; once you type something like 300 / 100 into such a calculator, it just shows you a 3, and then after you have executed another computation, that 3 is more or less permanently gone.
To do something with your data, you need to do one of two things: either store it into your environment somehow, or to "pipe" your data directly into a useful function.
In your question, you used this script:
1
3
2
7
6
summary()
I don't think it's possible to repair this strategy in the way that you're hoping -- and if it is possible, it's not quite the "right" approach. By typing the numbers on individual lines, you've structured them so that they'll evaluate individually and then evaporate. In order to run the summary() function on those numbers, you will need to bind them together inside a single vector somehow, then feed that vector into summary(). The "store it" approach would be
my_vector <- c(1, 3, 7, 2, 6)
summary(my_vector)
The importance isn't actually the parentheses; it's the function c(), which stands for concatenate, and instructs R to treat those 5 numbers as a collective object (i.e. a vector). We then pass that single object into my_vector().
To use the "piping" approach and avoid having to store something in the environment, you can do this instead (requires R 4.1.0+):
c(1, 3, 7, 2, 6) |> summary()
Note again that the use of c() is required, because we need to bind the five numbers together first. If you have an older version of R, you can get a slightly different pipe operator from the magrittr library instead that will work the same way. The point is that this "binding" part of the process is an essential part that can't be skipped.
Now, the crux of your question: presumably, your data doesn't really look like the example you used. Most likely, it's in some separate .csv file or something like that; if not, hopefully it is easy to get it into that format. Assuming this is true, this means that R will actually be able to do the heavy lifting for you in terms of formatting your data.
As a very simple example, let's say I have a plain text file, my_example.txt, whose contents are
1
3
7
2
6
In this case, I can ask R to parse this file for me. Assuming you're using RStudio, the simplest way to do this is to use the File -> Import Dataset part of the GUI. There are various options dealing with things such as headers, separators, and so forth, but I can't say much meaningful about what you'd need to do there without seeing your actual dataset.
When I import that file, I notice that it does two things in my R console:
my_example <- read.table(...)
View(my_example)
The first line stores an object (called a "data frame" in this case) in my environment; the second shows a nice view of how it's rendered. To get the summary I wanted, I just need to extract the vector of numbers I want, which I see from the view is called V1, which I can do with summary(my_example$V1).
This example is probably not helpful for your actual data set, because there are so many variations on the theme here, but the theme itself is important: point R at a file, as it to render an object, then work with that object. That's the approach I'd recommend instead of typing data as lines within an R script, as it's much faster and less error-prone.
Hopefully this will get you pointed in the right direction in terms of getting your data into R and working with it.
bmi<-function(x,y){
(x)/((y/100)^2)
}
bmi(70,177) it can work
but with apply() it does't work
apply(Student,1:2,bmi(Student$weight,Student$height))
Error in match.fun(FUN) :
'bmi(Student$weight, Student$height)' is not a function, character or symbol
It's a bit unclear what the goal is. If it's just to get an answer, then the comments do answer it. If on the other hand, the goal is to understand what you are doing wrong, then read on. I'd say the first error going from left to right is passing the whole dataframe. I would have only passed the 'height' and 'weight' columns.
The next error, again going from left to right, is the use of 1:2 as the second argument to apply. You obviously want to do this "by rows" which mean you should use only 1, i.e. the first dimension of the dataframe.
And the third error is using a function call rather than the function name. Functions with arguments in parentheses don't work when an R function (meaning apply in this case) is expecting a function name or an anonymous function as illustrated in comments.
Fourth error is not assigning the value to a column in your dataframe. So this probably would have succeeded in making the desired extra column via the apply method. But, as noted in comments this is not the most efficient method.:
Student$bmi_val <- apply(Student[ ,c("weight", "height")], bmi)
# didn't want my column name to be the same as the function name
The apply function was actually designed to work with matrices and arrays, so for many purposes it is ill-suited when used with dataframes. In this case where all the arguments to the bmi function are numeric and you can control the order of argument in the first argument to match the x and y positions, it's arguably an acceptable strategy, but not most R-ish method. When working with dates or factor variables, you should definitely avoid apply.
My goal of this code is to create a loop that aggregates each company's word frequency by a certain principle vector I created and adds it to a list. The problem is, after I run this, it only prints the 7 principles that I have rather than the word frequencies along side them. The word frequencies being the certain column of the FREQBYPRINC.AG data frame. Individually, running this code without the loop and just testing out a certain column, it works no problem. For some reason, the loop doesn't want to give me the correct data frames for the list. Any suggestions?
list.agg<-vector("list",ncol(FREQBYPRINC.AG)-2)
for (i in 1:14){
attach(FREQBYPRINC.AG)
list.agg[i]<-aggregate(FREQBYPRINC.AG[,i+1],by=list(Type=principle),FUN=sum,na.rm=TRUE)
}
I really wish I could help. After reading your statement, It seems that to you , you feel that the code should be working and it is not. Well maybe there exists a glitch.
Since you had previously specified list. agg as a list, you need to subset it with double square brackets. Try this one out:
list.agg<-vector("list",ncol(FREQBYPRINC.AG)-2)
for (i in 1:14){
list.agg[[i]]<-aggregate(FREQBYPRINC.AG[,i+1],by=list
(Type=principle),FUN=sum,na.rm=TRUE)}
I am putting together an R function that takes some undefined input through the ... argument described in the docs as:
"..." the special variable length argument ***
The idea is that the user will enter a number of column names here, each belonging to a dataset also specified by the user. These columns will then be cross-tabulated in comparison to the dependent variable by tapply. The function is to return a table (independent variable x indedependent variable).
Thus, I tried:
plotter=function(dataset, dependent_variable, ...)
{
indi_variables=list(...); # making a list of the ... input as described in the docs
result=with (dataset, tapply(dependent_variable, indi_variables, mean); # this fails
}
I figured this should work as tapply can take a list as input.
But it does not in this case ('Error in tapply...arguments must have same length') and I think it is because indi_variables is a list of strings.
If I input the contents of the list by hand and leave out the quotation marks, everything works just fine.
However, if the user feeds the function the column names as non-strings, R will interpret them as variable names; and I cannot figure out how to transform the list indi_variables in the right way, unsuccessfully trying things like this:
indi_variables=lapply(indi_variables, as.factor)
So I am wondering
What causes the error described above? Is my interpretation correct?
How would one go about transforming the list created through ... in the right way?
Is there an overall better way of doing this, in the input or the implementation of tapply?
Any help is much appreciated!
Thanks to Joran's helpful reading, I have come up with these improvements than make things work out...
indi_variables=substitute(list(...));
result=with (dataset, tapply(dependent_variable, eval(indi_variables, dataset), FUN=mean));
I am trying to use the write.table() function in R to write an output. It has 3 columns, first has numbers, the second has a color indicator (#000000) and the third one has a number again. Simple enough right?
Well, when I run my script step by step, the output is the desired one, something like this:
Tip ID Color Track Cluster ID
13298267 #FF0000 1
But when I try to do it through command line, with exactly the same number and the same arguments, I get something like this:
13298267 #FF0000 156
I tried changing the encoding and nothing happened, I tried changing the third column as a number instead of an integer. Nothing
Does someone have any idea of what might be happening?
This often happens if the class of the object is a factor, and not the expected numeric. This is because the print() functions for factors and numerics will both output an (often different) number at a casual glance.
Best way to avoid it is to always check using str() what classes your objects are.