R code for whatsapp average word length per person - r

I am new to R. Currently, I have parsed messages from a Whatsapp chat group and now I am trying to visualize data for average word length per member.
I am using this code to calculate the number of words for every time "Eddy" message
for(i in grep("Eddy",chatcsv[,2],fixed=TRUE)){
length(which(!is.na(chatcsv[i,4:111])))
}
This does not return any output or any error message.
My intention is to then sum up the total length and then divide by the number of times a person message. Lastly, I plan to place the average as a vector and visualize it as a bar graph.
Thank you

Your syntax is wrong. You should use:
allnames <- chatcsv[,2] #or cimilar
eddyindexes <- grep("Eddy",allnames,fixed=TRUE) #return indexes of eddys chats
eddyschats <- chatcsv[eddyindexes, 4:100]
eddysavgcharacters <- apply[eddyschats,function(x) mean(nchar(x))] #average nchars of eddys chats

I'm thinking you are coming from a non-functional language. (Not a language that is dysfunctional, but rather one that is not a "functional language".) Your expression length(which(!is.na(chatcsv[i,4:111]))) would do nothing, because it is inside a for loop but was not assigned to any name. It just disappears. You would have needed to create a named vector (let's say res) with res <-numeric(0) before your loop and then within your loop done:
res[i] <- length(which(!is.na(chatcsv[i,4:111])))
The earlier answerer was confusing grep and grepl in his comment. The grep function returns integer values; the grepl function returns logical vectors. They can both be used for indexing.
Whether that expression would give you the basis for furhter efforts is no clear. It would depend on the contents of chatcsv[i,4:111]. If the contents are single words then perhaps it would succeed. If they are sentences then it would not. The length function would just return the number of non-NA values in the row-vector. Only if your prior (undescribed) operations had created a clean set of "words" in that set of columns would you be getting meaningful results.

Related

R: Logical from 2 vectors on pattern match

Trying to clean up some dirty data (for work), my data frame has a column for customer information (for our example lets say store and product) in a long weird string, as well as a column for store and a column for product. I can parse the store and the product from the string. Here is where I arrive at my problem.
let's say (consider these vectors part of a larger dataframe, appended with data$ if that helps, I was just working with them as vectors thinking it may speed up the code not having to pull the whole dataframe):
WeirdString <- c("fname: john; lname:smith; store:Amazon Inc.; product:Echo", "fname: cindy; lname:smith; store:BestBuy; product:Ps-4","fname: jon; lname:smith; store:WALMART; product:Pants")
so I parse this to be:
WS_Store <- c("Amazon Inc.", "BestBuy", "WALMART")
WS_Prod <- c("Echo", "Ps-4", "Pants")
What's in the tables (i.e. the non-parsed columns) is:
DB_Store <- c("Amazon", "BEST BUY", "Other")
DB_Prod <- c("ECHO", "PS4", "Jeans")
I currently am using a for loop to loop through i to grepl the "true" string from the parsed string. This takes forever, and I know R was designed to use vectorized code, So my question is, how do I eliminate the loop and use something like lapply (which I tried, and failed at, because I'm not savvy enough with lapply), or some other vectorized thing?
My current code:
for(i in 1:nrow(data)){ # could be i in length(DB_prod) or whatever, all vectors are the same length)
Diff_Store[i] <- !grepl(DB_Store[i], WS_Store[i], ignore.case=T)
Diff_Prod[i] <- !grepl(DB_Prod[i] , WS_Prod[i] , ignore.case=T)
}
I intend to append those columns back into the dataframe, as the true goal is to diagnose why the database has this problem.
If there's a better way than this, rather than trying to vectorize it, I'm open to it. The data in the DB_Store is restricted to a specific number of "stores" (in the table it comes from) but in the string, it seems to be open, which is why I use the DB as the pattern, not the x. Product is similar, but not as restricted, this is why some have dashes and some don't. I would love to match "close things" like Ps-4 vs. PS4, but I will probably just build a table of matches once I see how weird the string gets. To be true though, the string may not match, which is represented by the Pants/Jeans thing. The dataset is 2.5 million records, and there are many different "stores" and "products", and I do want to make sure they match on the same line, not "is it in the database" (which is what previous questions seem to ask, can I see if a string is in a list of strings, rather than a 1:1 comparison, and the last question did end in a loop, which takes minutes and hours to run)
Thanks!
Please check if this works for you:
check <- function(vec_a, vec_b){
mat <- cbind(vec_a, vec_b)
diff <- apply(mat, 1, function(x) !grepl(pattern = x[1], x = x[2], ignore.case = TRUE))
diff
}
Use your different vectors for stores (or products) in the arguments vec_a and vec_b, respectively (example: diff_stores <- check(DB_Store, WS_Store) ). This function will return a logical vector with TRUE values referring to items that weren't a match in the two original vectors. Is this what you wanted?

R add to a list in a loop, using conditions

I have a data.frame dim = (200,500)
I want to do a shaprio.test on each column of my dataframe and append to a list. This is what I'm trying:
colstoremove <- list();
for (i in range(dim(I.df.nocov)[2])) {
x <- shapiro.test(I.df.nocov[1:200,i])
colstoremove[[i]] <- x[2]
}
However this is failing. Some pointers? (background is mainly python, not much of an R user)
Consider lapply() as any data frame passed into it runs operations on columns and the returned list will be equal to number of columns:
colstoremove <- lapply(I.df.noconv, function(col) shapiro.test(col)[2])
Here is what happens in
for (i in range(dim(I.df.nocov)[2]))
For the sake of example, I assume that I.df.nocov contains 100 rows and 5 columns.
dim(I.df.nocov) is the vector of I.df.nocov dimensions, i.e. c(100, 5)
dim(I.df.nocov)[2] is the 2nd dimension of I.df.nocov, i.e. 5
range(x)is a 2-element vector which contains minimal and maximal values of x. For example, range(c(4,10,1)) is c(1,10). So range(dim(I.df.nocov)[2]) is c(5,5).
Therefore, the loop iterate twice: first time with i=5, and second time also with i=5. Not surprising that it fails!
The problem is that R's function range and Python's function with the same name do completely different things. The equivalent of Python's range is called seq. For example, seq(5)=c(1,2,3,4,5), while seq(3,5)=c(3,4,5), and seq(1,10,2)=c(1,3,5,7,9). You may also write 1:n, it is the same as seq(n), and m:n is same as seq(m,n) (but the priority of ':' is very high, so 1:2*x is interpreted as (1:2)*x.
Generally, if something does not work in R, you should print the subexpressions from the innerwise to the outerwise. If some subexpression is too big to be printed, use str(x) (str means "structure"). And never assume that functions in Python and R are same! If there is a function with same name, it usually does a different thing.
On a side note, instead of dim(I.df.nocov)[2] you could just write ncol(I.df.nocov) (there is also a function nrow).

Loop and clear the basic function in R

I've got this dataset
install.packages("combinat")
install.packages("quantmod")
library(quantmod)
library(combinat)
library(utils)
getSymbols("AAPL",from="2012-01-01")
data<-AAPL
p1<-4
dO<-data[,1]
dC<-data[,4]
emaO<-EMA(dO,n=p1)
emaC<-EMA(dC,n=p1)
Pos_emaO_dO_UP<-emaO>dO
Pos_emaO_dO_D<-emaO<dO
Pos_emaC_dC_UP<-emaC>dC
Pos_emaC_dC_D<-emaC<dC
Pos_emaC_dO_D<-emaC<dO
Pos_emaC_dO_UP<-emaC>dO
Pos_emaO_dC_UP<-emaO>dC
Pos_emaO_dC_D<-emaO<dC
Profit_L_1<-((lag(dC,-1)-lag(dO,-1))/(lag(dO,-1)))*100
Profit_L_2<-(((lag(dC,-2)-lag(dO,-1))/(lag(dO,-1)))*100)/2
Profit_L_3<-(((lag(dC,-3)-lag(dO,-1))/(lag(dO,-1)))*100)/3
Profit_L_4<-(((lag(dC,-4)-lag(dO,-1))/(lag(dO,-1)))*100)/4
Profit_L_5<-(((lag(dC,-5)-lag(dO,-1))/(lag(dO,-1)))*100)/5
Profit_L_6<-(((lag(dC,-6)-lag(dO,-1))/(lag(dO,-1)))*100)/6
Profit_L_7<-(((lag(dC,-7)-lag(dO,-1))/(lag(dO,-1)))*100)/7
Profit_L_8<-(((lag(dC,-8)-lag(dO,-1))/(lag(dO,-1)))*100)/8
Profit_L_9<-(((lag(dC,-9)-lag(dO,-1))/(lag(dO,-1)))*100)/9
Profit_L_10<-(((lag(dC,-10)-lag(dO,-1))/(lag(dO,-1)))*100)/10
which are given to this frame
frame<-data.frame(Pos_emaO_dO_UP,Pos_emaO_dO_D,Pos_emaC_dC_UP,Pos_emaC_dC_D,Pos_emaC_dO_D,Pos_emaC_dO_UP,Pos_emaO_dC_UP,Pos_emaO_dC_D,Profit_L_1,Profit_L_2,Profit_L_3,Profit_L_4,Profit_L_5,Profit_L_6,Profit_L_7,Profit_L_8,Profit_L_9,Profit_L_10)
colnames(frame)<-c("Pos_emaO_dO_UP","Pos_emaO_dO_D","Pos_emaC_dC_UP","Pos_emaC_dC_D","Pos_emaC_dO_D","Pos_emaC_dO_UP","Pos_emaO_dC_UP","Pos_emaO_dC_D","Profit_L_1","Profit_L_2","Profit_L_3","Profit_L_4","Profit_L_5","Profit_L_6","Profit_L_7","Profit_L_8","Profit_L_9","Profit_L_10")
There is vector with variables for later usage
vector<-c("Pos_emaO_dO_UP","Pos_emaO_dO_D","Pos_emaC_dC_UP","Pos_emaC_dC_D","Pos_emaC_dO_D","Pos_emaC_dO_UP","Pos_emaO_dC_UP","Pos_emaO_dC_D")
I made all possible combination with 4 variables of the vector (there are no depended variables)
comb<-as.data.frame(combn(vector,4))
comb
and get out the ,,nonsense" combination (where are both possible values of variable)
rc<-comb[!sapply(comb, function(x) any(duplicated(sub('_D|_UP', '', x))))]
rc
Then I prepare the first combination to later subseting
var<-paste(rc[,1],collapse=" & ")
var
and subset the frame (with all DVs)
kr<-eval(parse(text=paste0('subset(frame,' , var,')' )))
kr
Now I have the subseted df by the first combination of 4 variables.
Then I used the evaluation function on it
evaluation<-function(x){
s_1<-nrow(x[x$Profit_L_1>0,])/nrow(x)
s_2<-nrow(x[x$Profit_L_2>0,])/nrow(x)
s_3<-nrow(x[x$Profit_L_3>0,])/nrow(x)
s_4<-nrow(x[x$Profit_L_4>0,])/nrow(x)
s_5<-nrow(x[x$Profit_L_5>0,])/nrow(x)
s_6<-nrow(x[x$Profit_L_6>0,])/nrow(x)
s_7<-nrow(x[x$Profit_L_7>0,])/nrow(x)
s_8<-nrow(x[x$Profit_L_8>0,])/nrow(x)
s_9<-nrow(x[x$Profit_L_9>0,])/nrow(x)
s_10<-nrow(x[x$Profit_L_10>0,])/nrow(x)
n_1<-nrow(x[x$Profit_L_1>0,])/nrow(frame)
n_2<-nrow(x[x$Profit_L_2>0,])/nrow(frame)
n_3<-nrow(x[x$Profit_L_3>0,])/nrow(frame)
n_4<-nrow(x[x$Profit_L_4>0,])/nrow(frame)
n_5<-nrow(x[x$Profit_L_5>0,])/nrow(frame)
n_6<-nrow(x[x$Profit_L_6>0,])/nrow(frame)
n_7<-nrow(x[x$Profit_L_7>0,])/nrow(frame)
n_8<-nrow(x[x$Profit_L_8>0,])/nrow(frame)
n_9<-nrow(x[x$Profit_L_9>0,])/nrow(frame)
n_10<-nrow(x[x$Profit_L_10>0,])/nrow(frame)
pr_1<-sum(kr[,"Profit_L_1"])/nrow(kr[,kr=="Profit_L_1"])
pr_2<-sum(kr[,"Profit_L_2"])/nrow(kr[,kr=="Profit_L_2"])
pr_3<-sum(kr[,"Profit_L_3"])/nrow(kr[,kr=="Profit_L_3"])
pr_4<-sum(kr[,"Profit_L_4"])/nrow(kr[,kr=="Profit_L_4"])
pr_5<-sum(kr[,"Profit_L_5"])/nrow(kr[,kr=="Profit_L_5"])
pr_6<-sum(kr[,"Profit_L_6"])/nrow(kr[,kr=="Profit_L_6"])
pr_7<-sum(kr[,"Profit_L_7"])/nrow(kr[,kr=="Profit_L_7"])
pr_8<-sum(kr[,"Profit_L_8"])/nrow(kr[,kr=="Profit_L_8"])
pr_9<-sum(kr[,"Profit_L_9"])/nrow(kr[,kr=="Profit_L_9"])
pr_10<-sum(kr[,"Profit_L_10"])/nrow(kr[,kr=="Profit_L_10"])
mat<-matrix(c(s_1,n_1,pr_1,s_2,n_2,pr_2,s_3,n_3,pr_3,s_4,n_4,pr_4,s_5,n_5,pr_5,s_6,n_6,pr_6,s_7,n_7,pr_7,s_8,n_8,pr_8,s_9,n_9,pr_9,s_10,n_10,pr_10),ncol=3,nrow=10,dimnames=list(c(1:10),c("s","n","pr")))
df<-as.data.frame(mat)
return(df)
}
result<-evaluation(kr)
result
And I need to help in several cases.
1, in evaluation function the way the matrix is made is wrong (s_1,n_1,pr_1 are starting in first column but I need to start the order by rows)
2, I need to use some loop/lapply function to go trough all possible combinations (not only the first one like in this case (var<-paste(rc[,1],collapse=" & ")) and have the understandable output where is evaluation function used on every combination and I will be able to see for which combination of variables is the evaluation done (understand I need to recognize for what is this evaluation made) and compare evaluation results for each combination.
3, This is not main point, BUT I generally want to evaluate all possible combinations (it means for 2:n number of variables and also all combinations in each of them) and then get the best possible combination according to specific DV (Profit_L_1 or Profit_L_2 and so on). And I am so weak in looping now, so, if it this possible, keep in mind what am I going to do with it later.
Thanks, feel free to update, repair or improve the question (if there is something which could be done way more easily, effectively - do it - I am open for every senseful advice.

Find same observations of a list/matrix in another matrix

For a marketing class I have to write a function that calculates the retention rate of the customers (probability that a customer still is a customer). I've come so far that I isolated the ids of the individual customers and stored them in the matrix first.transactions.data. I then split them into cohorts (group of customers by time) with split() and stored them in the list cohort.
Now comes my problem: I calculated another sub-matrix from the full data set called final.period.data where I will calculate the retention rate. However, therefore I have to isolate the ids in final.period.data for each cohort. My instructor told me that I should create an additional column in final.period.data that shows TRUE or FALSE depending on whether the cohort's id and final.period.data's id are the same. For this I tried to use exists, but I always receive error messages. I tried the following:
final.period.data <- if(exists(cohort$'1'$id, where = final.period.data$id) final.period.data$same = TRUE)
but always receive error messages such as: unexpected symbol or invalid first argument. I also tried to convert the list cohort into a matrix but this didn't help either. How do I have to change the exist command or is there a simpler way to locate cohort's ids in final.period.data?
Thank you for your help.
You can just create a function that does what you want:
funct <-(final.period.data){
if (final.period.data$cohort =='1' & final.period.data$id ==<condition2>){
#Change the number for the TRUE condition}
else{ #If it doesn't fit the two conditions
#Change the number for the FALSE condition}
}
vector <- c(nrow(final.period.data))
final.period.data <- cbind(vector)
And use it as the apply function. Here will you find more information about apply
But I usually do it with a for loop, first creating the new column and then adding it to the data frame.

Efficient function to return varying length vector from lookup table

I have three data sources:
types<-c(1,3,3)
places<-list(c(1,2,3),1,c(2,3))
lookup.counts<-as.data.frame(matrix(runif(9,min=0,max=10),nrow=3,ncol=3))
assigned.places<-rep.int(0,length(types))
the numbers in the "types" vector tell me what 'type' a given observation is. The vectors in the places list tell me which places the observation can be found in (some observations are found in only one place, others in all places). By definition there is one entry in types and one list in places for each observation. Lookup.counts tells me how many observations of each type are located in each place (generated from another data source).
I want to randomly assign each observation to a place based on a probability generated from lookup.counts. Using for loops it looks something like"
for (i in 1:length(types)){
row<-types[i]
columns<-places[[i]]
this.obs<-lookup.counts[row,columns] #the counts of this type in each place
total<-sum(this.obs)
this.obs<-this.obs/total #the share of observations of this type in these places
pick<-runif(1,min=0,max=1)
#the following should really be a 'while' loop, but regardless it needs help
for(j in 1:length(this.obs[])){
if(this.obs[j] > pick){
#pick is less than this county so assign
pick<- 100 #just a way of making sure an observation doesn't get assigned twice
assigned.places[i]<-colnames(lookup.counts)[j]
}else{
#pick is greater, move to the next category
pick<- pick-this.obs[j]
}
}
}
I have been trying to vectorize this somehow, but am getting hung up on the variable length of 'places' and of 'this.obs'
In practice, of course, the lookup.counts table is quite a bit bigger (500 x 40) and I have some 900K observations with places lists of length 1 through length 39.
To vectorize the inner loop, you can use sample or sample.int to choose from several alternaives with prescribed probabilities. Unless I read your code incorrectly, you want something like this:
assigned.places[i] <- sample(colnames(this.obs), 1, prob = this.obs)
I'm a bit surprised that you're using colnames(lookup.counts) instead. Shouldn't this be subset by columns as well? It seems that either I missed something, or there is a bug in your code.
the different lengths of your lists are a severe obstacle to vectorizing your outer loops. Perhaps you could use the Matrix package to store that information as sparse matrices. Then you could simply multiply probabilities by that vector to exclude those columns which are not in the places list of a given observation. But as you'd probably still use apply for the above sampling code, you might as well keep the list and use some form of apply to iterate over that.
The overall result might look somewhat like this:
assigned.places <- colnames(lookup.counts)[
apply(cbind(types, places), 1, function(x) {
sample(x[[2]], 1, prob=lookup.counts[x[[1]],x[[2]]])
})
]
The use of cbind and apply isn't particularly beautiful, but seems to work. Each x is a list of two items, x[[1]] being the type and x[[2]] being the corresponding places. We use these to index lookup.counts just as you did. Then we use the found counts as relative probabilities when choosing the index of one of the columns we used in the subscript. Only after all these numbers have been assembled into a single vector by apply will the indices be turned into names based on colnames.
You can check whether things are faster if you don't cbindstuff together, but instead iterate over the indices only:
assigned.places <- colnames(lookup.counts)[
sapply(1:length(types), function(i) {
sample(places[[i]], 1, prob=lookup.counts[types[i],places[[i]]])
})
]
This appears to work as well:
# More convenient if lookup.counts is a matrix.
lookup.counts<-matrix(runif(9,min=0,max=10),nrow=3,ncol=3)
colnames(lookup.counts)<-paste0('V',1:ncol(lookup.counts))
# A function that does what the for loop does for each i
test<-function(i) {
this.places<-colnames(lookup.counts)[places[[i]]]
this.obs<-lookup.counts[types[i],this.places]
sample(this.places,size=1,prob=this.obs)
}
# Applies the function for all i
sapply(1:length(types),test)

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