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)
Related
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?
I'm currently reading "Practical Statistics for Data Scientists" and following along in R as they demonstrate some code. There is one chunk of code I'm particularly struggling to follow the logic of and was hoping someone could help. The code in question is creating a dataframe with 1000 rows where each observation is the mean of 5 randomly drawn income values from the dataframe loans_income. However, I'm getting confused about the logic of the code as it is fairly complicated with a tapply() function and nested rep() statements.
The code to create the dataframe in question is as follows:
samp_mean_5 <- data.frame(income = tapply(sample(loans_income$income,1000*5),
rep(1:1000,rep(5,1000)),
FUN = mean),
type='mean_of_5')
In particular, I'm confused about the nested rep() statements and the 1000*5 portion of the sample() function. Any help understanding the logic of the code would be greatly appreciated!
For reference, the original dataset loans_income simply has a single column of 50,000 income values.
You have 50,000 loans_income in a single vector. Let's break your code down:
tapply(sample(loans_income$income,1000*5),
rep(1:1000,rep(5,1000)),
FUN = mean)
I will replace 1000 with 10 and income with random numbers, so it's easier to explain. I also set set.seed(1) so the result can be reproduced.
sample(loans_income$income,1000*5)
We 50 random incomes from your vector without replacement. They are (temporarily) put into a vector of length 50, so the output looks like this:
> sample(runif(50000),10*5)
[1] 0.73283101 0.60329970 0.29871173 0.12637654 0.48434952 0.01058067 0.32337850
[8] 0.46873561 0.72334215 0.88515494 0.44036341 0.81386225 0.38118213 0.80978822
[15] 0.38291273 0.79795343 0.23622492 0.21318431 0.59325586 0.78340477 0.25623138
[22] 0.64621658 0.80041393 0.68511759 0.21880083 0.77455662 0.05307712 0.60320912
[29] 0.13191926 0.20816298 0.71600799 0.70328349 0.44408218 0.32696205 0.67845445
[36] 0.64438336 0.13241312 0.86589561 0.01109727 0.52627095 0.39207860 0.54643661
[43] 0.57137320 0.52743012 0.96631114 0.47151170 0.84099503 0.16511902 0.07546454
[50] 0.85970500
rep(1:1000,rep(5,1000))
Now we are creating an indexing vector of length 50:
> rep(1:10,rep(5,10))
[1] 1 1 1 1 1 2 2 2 2 2 3 3 3 3 3 4 4 4 4 4 5 5 5 5 5 6 6 6
[29] 6 6 7 7 7 7 7 8 8 8 8 8 9 9 9 9 9 10 10 10 10 10
Those indices "group" the samples from step 1. So basically this vector tells R that the first 5 entries of your "sample vector" belong together (index 1), the next 5 entries belong together (index 2) and so on.
FUN = mean
Just apply the mean-function on the data.
tapply
So tapply takes the sampled data (sample-part) and groups them by the second argument (the rep()-part) and applies the mean-function on each group.
If you are familiar with data.frames and the dplyr package, take a look at this (only the first 10 rows are displayed):
set.seed(1)
df <- data.frame(income=sample(runif(5000),10*5), index=rep(1:10,rep(5,10)))
income index
1 0.42585569 1
2 0.16931091 1
3 0.48127444 1
4 0.68357403 1
5 0.99374923 1
6 0.53227877 2
7 0.07109499 2
8 0.20754511 2
9 0.35839481 2
10 0.95615917 2
I attached the an index to the random numbers (your income). Now we calculate the mean per group:
df %>%
group_by(index) %>%
summarise(mean=mean(income))
which gives us
# A tibble: 10 x 2
index mean
<int> <dbl>
1 1 0.551
2 2 0.425
3 3 0.827
4 4 0.391
5 5 0.590
6 6 0.373
7 7 0.514
8 8 0.451
9 9 0.566
10 10 0.435
Compare it to
set.seed(1)
tapply(sample(runif(5000),10*5),
rep(1:10,rep(5,10)),
mean)
which yields basically the same result:
1 2 3 4 5 6 7 8 9
0.5507529 0.4250946 0.8273149 0.3905850 0.5902823 0.3730092 0.5143829 0.4512932 0.5658460
10
0.4352546
Probably a similar situation has already been solved but I could not find it.
I have a mapper data frame like the following
mapper
bucket_label bucket_no
1 (-Inf; 9.99) 1
2 (25.01; 29.99) 1
3 (29.99; 30.01) 1
4 (30.01; Inf) 1
5 (19.99; 20.01) 2
6 (20.01; 24.99) 2
7 (24.99; 25.01) 2
8 (9.99; 10.11) 3
9 (10.11; 14.99) 3
10 (14.99; 15.01) 3
11 (15.01; 19.99) 3
and a vector x with random data
x <- rnorm(100)*100
I need to set the corresponding bucket for each entry of this in a quick way and findInterval and cut seem not to help for this issue.
Say I have data that look like this:
level start end
1 1 133.631 825.141
2 2 133.631 155.953
3 3 146.844 155.953
4 2 293.754 302.196
5 3 293.754 302.196
6 4 293.754 301.428
7 2 326.253 343.436
8 3 326.253 343.436
9 4 333.827 343.436
10 2 578.066 611.766
11 3 578.066 611.766
12 4 578.066 587.876
13 4 598.052 611.766
14 2 811.228 825.141
15 3 811.228 825.141
or this:
level start end
1 1 3.60353 1112.62000
2 2 3.60353 20.35330
3 3 3.60353 8.77526
4 2 72.03720 143.60700
5 3 73.50530 101.13200
6 4 73.50530 81.64660
7 4 92.19030 101.13200
8 3 121.28500 143.60700
9 4 121.28500 128.25900
10 2 167.19700 185.04800
11 3 167.19700 183.44600
12 4 167.19700 182.84600
13 2 398.12300 418.64300
14 3 398.12300 418.64300
15 2 445.83600 454.54500
16 2 776.59400 798.34800
17 3 776.59400 796.64700
18 4 776.59400 795.91300
19 2 906.68800 915.89700
20 3 906.68800 915.89700
21 2 1099.44000 1112.62000
22 3 1099.44000 1112.62000
23 4 1100.14000 1112.62000
They produce the following graphs:
As you can see there are several time intervals at different levels. The level-1 interval always spans the entire duration of the time of interest. Levels 2+ have time intervals that are shorter.
What I would like to do is select the maximum number of non-overlapping time intervals covering each period that contain the maximum number of total time within them. I have marked in pink which ones those would be.
For small dataframes it is possible to brute force this, but obviously there should be some more logical way of doing this. I'm interested in hearing some ideas about what I should try.
EDIT:
I think one thing that could help here is the column 'level'. The results come from Kleinberg's burst detection algorithm (package 'bursts'). You will note that the levels are hierarchically organized. Levels of the same number cannot overlap. However levels successively increasing e.g. 2,3,4 in successive rows can overlap.
In essence, I think the problem could be shortened to this. Take the levels produced, but remove level 1. This would be the vector for the 2nd example:
2 3 2 3 4 4 3 4 2 3 4 2 3 2 2 3 4 2 3 2 3 4
Then, look at the 2s... if there are fewer than or only one '3' then that 2 is the longest interval. But if there are two or more 3's between successive 2's, then those 3s should be counted. Do this iteratively for each level. I think that should work...?
e.g.
vec<-df$level %>% as.vector() %>% .[-1]
vec
#[1] 2 3 2 3 4 4 3 4 2 3 4 2 3 2 2 3 4 2 3 2 3 4
max(vec) #4
vec3<-vec #need to find two or more 4's between 3s
vec3[vec3==3]<-NA
names(vec3)<-cumsum(is.na(vec3))
0 1 1 2 2 2 3 3 3 4 4 4 5 5 5 6 6 6 7 7 8 8
2 NA 2 NA 4 4 NA 4 2 NA 4 2 NA 2 2 NA 4 2 NA 2 NA 4
vec3.res<-which(table(vec3,names(vec3))["4",]>1)
which(names(vec3)==names(vec3.res) & vec3==4) #5 6
The above identifies rows 5 and 6 (which equate to rows 6 and 7 in original df) as having two fours that lie between 3's. Perhaps something using this sort of approach might work?
OK here is a stab using your second data set to test. This might not be correct in all cases!!
library(data.table)
dat <- fread("data.csv")
dat[,use:="maybe"]
make.pass <- function(dat,low,high,the.level,use) {
check <- dat[(use!="no" & level > the.level)]
check[,contained.by.above:=(low<=start & end<=high)]
check[,consecutive.contained.by.above:=
(contained.by.above &
!is.na(shift(contained.by.above,1)) &
shift(contained.by.above,1)),by=level]
if(!any(check[,consecutive.contained.by.above])) {
#Cause a side effect where we've learned we don't care:
dat[check[(contained.by.above),rownum],use:="no"]
print(check)
return("yes")
} else {
return("no")
}
}
dat[,rownum:=.I]
dat[level==1,use:=make.pass(dat,start,end,level,use),by=rownum]
dat
dat[use=="maybe" & level==2,use:=make.pass(dat,start,end,level,use),by=rownum]
dat
dat[use=="maybe" & level==3,use:=make.pass(dat,start,end,level,use),by=rownum]
dat
#Finally correct for last level
dat[use=="maybe" & level==4,use:="yes"]
I wrote these last steps out so you can trace in your own interactive session to see what's happening (see the print to get an idea) but you can remove the print and also condense the last steps into something like lapply(1:dat[,max(level)-1], function(the.level) dat[use=="maybe" & level==the.level,use:=make.pass......]) In response to your comment if there are an arbitrary number of levels you will definitely want to use this formalism, and follow it with a final call to dat[use=="maybe" & level==max(level),use:="yes"].
Output:
> dat
level start end use rownum
1: 1 3.60353 1112.62000 no 1
2: 2 3.60353 20.35330 yes 2
3: 3 3.60353 8.77526 no 3
4: 2 72.03720 143.60700 no 4
5: 3 73.50530 101.13200 no 5
6: 4 73.50530 81.64660 yes 6
7: 4 92.19030 101.13200 yes 7
8: 3 121.28500 143.60700 yes 8
9: 4 121.28500 128.25900 no 9
10: 2 167.19700 185.04800 yes 10
11: 3 167.19700 183.44600 no 11
12: 4 167.19700 182.84600 no 12
13: 2 398.12300 418.64300 yes 13
14: 3 398.12300 418.64300 no 14
15: 2 445.83600 454.54500 yes 15
16: 2 776.59400 798.34800 yes 16
17: 3 776.59400 796.64700 no 17
18: 4 776.59400 795.91300 no 18
19: 2 906.68800 915.89700 yes 19
20: 3 906.68800 915.89700 no 20
21: 2 1099.44000 1112.62000 yes 21
22: 3 1099.44000 1112.62000 no 22
23: 4 1100.14000 1112.62000 no 23
level start end use rownum
On the off chance this is correct, the algorithm can roughly be described as follows:
Mark all the intervals as possible.
Start with a given level. Pick a particular interval (by=rownum) say called X. With X in mind, subset a copy of the data to all higher-level intervals.
Mark any of these that are contained in X as "contained in X".
If consecutive intervals at the same level are contained in X, X is no good b/c it wastes intervals. In this case label X's "use" variable as "no" so we'll never think about X again. [Note: if it's possible that non-consecutive intervals are contained in X, or that containing multiple intervals across levels could ruin X's viability, then this logic might need to be changed to count contained intervals instead of finding consecutive ones. I didn't think about this at all, but it's just occurring to me now, so use at your own risk.]
On the other hand, if X passed the test, then we've already established it's good. Mark it as a "yes." But importantly, we also have to mark any single interval contained in X as "no," or else when we iterate the step it will forget that it was contained inside a good interval and mark itself as "yes" as well. This is the side effect step.
Now, iterate, ignoring any results that we've already determined.
Finally any "maybe"s leftover at the highest level are automatically in.
Let me know what you think of this--this is a rough draft and some aspects might not be correct.
I always get angry at my R code when I have to process dataframes, i.e. filtering out certain rows. The code gets very illegible as I tend to choose meaningful, but long, names for my objects. An example:
all.mutations.extra.large.name <- read.delim(filename)
head(all.mutations.extra.large.name)
id gene pos aa consequence V
ENSG00000105732 ZN574_HUMAN 81 x/N missense_variant 3
ENSG00000125879 OTOR_HUMAN 7 V/3 missense_variant 2
ENSG00000129194 SOX15_HUMAN 20 N/T missense_variant 3
ENSG00000099204 ABLM1_HUMAN 33 H/R missense_variant 2
ENSG00000103335 PIEZ1_HUMAN 11 Q/R missense_variant 3
ENSG00000171533 MAP6_HUMAN 39 A/G missense_variant 3
all.mutations.extra.large.name <- all.mutations.extra.large.name[which(all.mutations.extra.large.name$gene == ZN574_HUMAN)]
So in order to kick out all other lines in which I am not interested I need to reference 3 times the object all.mutations.extra.large.name. And reating this kind of step for different columns makes the code really difficult to understand.
Therefore my question: Is there a way to filter out rows by a criterion without referencing the object 3 times. Something like this would be beautiful: myobj[,gene=="ZN574_HUMAN"]
You can use subset for that:
subset(all.mutations.extra.large.name, gene == "ZN574_HUMAN")
Several options:
all.mutations.extra.large.name <- data.frame(a=1:5, b=2:6)
within(all.mutations.extra.large.name, a[a < 3] <- 0)
a b
1 0 2
2 0 3
3 3 4
4 4 5
5 5 6
transform(all.mutations.extra.large.name, b = b^2)
a b
1 1 4
2 2 9
3 3 16
4 4 25
5 5 36
Also check ?attach if you would like to avoid repetitive typing like all.mutations.extra.large.name$foo.