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Say, I have a dataframe df in R as follows,
id inflam
1 1 0.03093764
2 2 0.50115406
3 3 0.82153770
4 4 0.01985961
5 5 0.04994588
6 6 0.91714810
7 7 0.83438400
8 8 0.80832225
9 9 0.12360681
10 10 0.08490079
I can access the entirety of the inflam column by indexing as df[,2] or df[2]. However, typeof(df[,2]) returns double, whereas typeof(df[2]) returns list. The comma seems to be the differentiator, but why is this the case? What is going on under the hood?
Suppose we have a vector, we can easily enough lapply, sapply or map across 1 element at a time.
Is there a way to do the same across groups of (>1) elements of the vector?
Example
Suppose we are constructing API calls by appending comma-separated user_identifiers to the URL, like so:
user_identifiers <- c("0011399", "0011400", "0013581", "0013769", "0013770", "0018374",
"0018376", "0018400", "0018401", "0018410", "0018415", "0018417",
"0018419", "0018774", "0018775", "0018776", "0018777", "0018778",
"0018779", "0021627", "0023492", "0023508", "0023511", "0023512",
"0024120", "0025672", "0025673", "0025675", "0025676", "0028226",
"0028227", "0028266", "0028509", "0028510", "0028512", "0028515",
"0028518", "0028520", "0028523", "0029160", "0033141", "0034586",
"0035035", "0035310", "0035835", "0035841", "0035862", "0036503",
"0036580", "0036583", "0036587", "0037577", "0038582", "0038583",
"0038587", "0039727", "0039729", "0039731", "0044703", "0044726"
)
get_data <- function(user_identifier) {
url <- paste0("https://www.myapi.com?userIdentifier=",
paste0(user_identifier, collapse=","))
fromJSON(url)
}
In the above, get_data(user_identifiers) would return the APIs response for all 60 user_identifiers in one single request.
But suppose the API accepts a maximum of 10 identifiers at a time (so we cannot do all 60 at once).
A simple solution could be to simply map/lapply/sapply over each element, e.g. sapply(get_data, user_identifiers - this would work fine - however, we would make 60 API calls, when all we really need is 6. If we could map/lapply/sapply over groups of 10 at a time; that would be ideal
Question
Is there an elegant way to map/lapply/sapply over groups of n elements at a time (where n>1)?
We can split user_identifiers in groups of 10 and use sapply/map/lapply
sapply(split(user_identifiers, gl(length(user_identifiers)/10, 10)), get_data)
where gl creates groups from 1 to 6 each of length 10.
gl(length(user_identifiers)/10, 10)
# [1] 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3
# 4 4 4 4 4 4 4 4 4 4 5 5 5 5 5 5 5 5 5 5 6 6 6 6 6 6 6 6 6 6
#Levels: 1 2 3 4 5 6
The same groups can be created with rep
rep(1:ceiling(length(user_identifiers)/10), each = 10)
As #thelatemail mentioned, we can use cut and specify number of groups to cut the data into
sapply(split(user_identifiers, cut(seq_along(user_identifiers),6)), get_data)
I'm new with R and I have the following problem. Maybe it's a really easy question, but I don't know the terms to search for an answer.
My problem:
I have several persons, each person is assigned a studynumber (SN). And each SN has one or more tests being performed, the test can have multiple results.
My data is long at the moment, but I need it to be wide (one row for each SN).
For example:
What I have:
SN testnumbers result
1 1 1234 6
2 1 1234 9
3 2 4567 6
4 3 5678 9
5 3 8790 9
What I want:
SN test1result1 test1result2 test2result1
1 1 6 6 NA
2 2 6 NA NA
3 3 9 NA 9
So I need to renumber the testnumbers into test 1 etc for each SN, in order to use the spread function, I think. But I don't know how.
I did manage to renumber testnumber into a list of 1 till the last unique testnumber, but still the wide dataframe looks awful.
I have a dataset of Ages for the customer and I wanted to make a frequency distribution by 9 years of a gap of age.
Ages=c(83,51,66,61,82,65,54,56,92,60,65,87,68,64,51,
70,75,66,74,68,44,55,78,69,98,67,82,77,79,62,38,88,76,99,
84,47,60,42,66,74,91,71,83,80,68,65,51,56,73,55)
My desired outcome would be similar to below-shared table, variable names can be differed(as you wish)
Could I use binCounts code into it ? if yes could you help me out using the code as not sure of bx and idxs in this code?
binCounts(x, idxs = NULL, bx, right = FALSE) ??
Age Count
38-46 3
47-55 7
56-64 7
65-73 14
74-82 10
83-91 6
92-100 3
Much Appreciated!
I don't know about the binCounts or even the package it is in but i have a bare r function:
data.frame(table(cut(Ages,0:7*9+37)))
Var1 Freq
1 (37,46] 3
2 (46,55] 7
3 (55,64] 7
4 (64,73] 14
5 (73,82] 10
6 (82,91] 6
7 (91,100] 3
To exactly duplicate your results:
lowerlimit=c(37,46,55,64,73,82,91,101)
Labels=paste(head(lowerlimit,-1)+1,lowerlimit[-1],sep="-")#I add one to have 38 47 etc
group=cut(Ages,lowerlimit,Labels)#Determine which group the ages belong to
tab=table(group)#Form a frequency table
as.data.frame(tab)# transform the table into a dataframe
group Freq
1 38-46 3
2 47-55 7
3 56-64 7
4 65-73 14
5 74-82 10
6 83-91 6
7 92-100 3
All this can be combined as:
data.frame(table(cut(Ages,s<-0:7*9+37,paste(head(s+1,-1),s[-1],sep="-"))))
I have binned data that looks like this:
(8.048,18.05] (-21.95,-11.95] (-31.95,-21.95] (18.05,28.05] (-41.95,-31.95]
81 76 18 18 12
(-132,-122] (-122,-112] (-112,-102] (-162,-152] (-102,-91.95]
6 6 6 5 5
(-91.95,-81.95] (-192,-182] (28.05,38.05] (38.05,48.05] (58.05,68.05]
5 4 4 4 4
(78.05,88.05] (98.05,108] (-562,-552] (-512,-502] (-482,-472]
4 4 3 3 3
(-452,-442] (-412,-402] (-282,-272] (-152,-142] (48.05,58.05]
3 3 3 3 3
(68.05,78.05] (118,128] (128,138] (-582,-572] (-552,-542]
3 3 3 2 2
(-532,-522] (-422,-412] (-392,-382] (-362,-352] (-262,-252]
2 2 2 2 2
(-252,-242] (-142,-132] (-81.95,-71.95] (148,158] (-1402,-1392]
2 2 2 2 1
(-1372,-1362] (-1342,-1332] (-942,-932] (-862,-852] (-822,-812]
1 1 1 1 1
(-712,-702] (-682,-672] (-672,-662] (-632,-622] (-542,-532]
1 1 1 1 1
(-502,-492] (-492,-482] (-472,-462] (-462,-452] (-442,-432]
1 1 1 1 1
(-432,-422] (-352,-342] (-332,-322] (-312,-302] (-302,-292]
1 1 1 1 1
(-202,-192] (-182,-172] (-172,-162] (-51.95,-41.95] (88.05,98.05]
1 1 1 1 1
(108,118] (158,168] (168,178] (178,188] (298,308]
1 1 1 1 1
(318,328] (328,338] (338,348] (368,378] (458,468]
1 1 1 1 1
How can I plot this data so that the bin is sorted from most negative on the left to most positive on the right? Currently my graph looks like this. Notice that it is not sorted at all. In particular the second bar (value = 76) is placed to the right of the first:
(8.048,18.05] (-21.95,-11.95]
81 76
This is the command I use to plot:
barplot(x,ylab="Number of Unique Tags", xlab="Expected - Observed")
I really want to help answer your question, but I gotta tell you, I can't make heads or tails of your data. I see a lot of opening parenthesis but no closing ones. The data looks sorted descending by whatever the values are on the bottom of each row. I have no idea what to make out of a value like "(8.048,18.05]"
Am I missing something obvious? Can you make a more simple example where your data structure is not a factor?
I would generally expect a data frame or a matrix with two columns, one for the X and one for the Y.
See if this example of sorting helps (I'm sort of shooting in the dark here)
tN <- table(Ni <- rpois(100, lambda=5))
r <- barplot(tN)
#stop here and examine the plot
#the next bit converts the matrix to a data frame,
# sorts it, and plots it again
df<-data.frame(tN)
df2<-df[order(df$Freq),]
barplot(df2$Freq)