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Suppose, I have a dataframe, df, and I want to create a new column called "c" based on the addition of two existing columns, "a" and "b". I would simply run the following code:
df$c <- df$a + df$b
But I also want to do this for many other columns. So why won't my code below work?
# Reproducible data:
martial_arts <- data.frame(gym_branch=c("downtown_a", "downtown_b", "uptown", "island"),
day_boxing=c(5,30,25,10),day_muaythai=c(34,18,20,30),
day_bjj=c(0,0,0,0),day_judo=c(10,0,5,0),
evening_boxing=c(50,45,32,40), evening_muaythai=c(50,50,45,50),
evening_bjj=c(60,60,55,40), evening_judo=c(25,15,30,0))
# Creating a list of the new column names of the columns that need to be added to the martial_arts dataframe:
pattern<-c("_boxing","_muaythai","_bjj","_judo")
d<- expand.grid(paste0("martial_arts$total",pattern))
# Creating lists of the columns that will be added to each other:
e<- names(martial_arts %>% select(day_boxing:day_judo))
f<- names(martial_arts %>% select(evening_boxing:evening_judo))
# Writing a function and using mapply:
kick_him <- function(d,e,f){d <- rowSums(martial_arts[ , c(e, f)], na.rm=T)}
mapply(kick_him,d,e,f)
Now, mapply produces the correct results in terms of the addition:
> mapply(ff,d,e,f)
Var1 <NA> <NA> <NA>
[1,] 55 84 60 35
[2,] 75 68 60 15
[3,] 57 65 55 35
[4,] 50 80 40 0
But it doesn't add the new columns to the martial_arts dataframe. The function in theory should do the following
martial_arts$total_boxing <- martial_arts$day_boxing + martial_arts$evening_boxing
...
...
martial_arts$total_judo <- martial_arts$day_judo + martial_arts$evening_judo
and add four new total columns to martial_arts.
So what am I doing wrong?
The assignment is wrong here i.e. instead of having martial_arts$total_boxing as a string, it should be "total_boxing" alone and this should be on the lhs of the Map/mapply. As the OP already created the 'martial_arts$' in 'd' dataset as a column, we are removing the prefix part and do the assignment
kick_him <- function(e,f){rowSums(martial_arts[ , c(e, f)], na.rm=TRUE)}
martial_arts[sub(".*\\$", "", d$Var1)] <- Map(kick_him, e, f)
-check the dataset now
> martial_arts
gym_branch day_boxing day_muaythai day_bjj day_judo evening_boxing evening_muaythai evening_bjj evening_judo total_boxing total_muaythai total_bjj total_judo
1 downtown_a 5 34 0 10 50 50 60 25 55 84 60 35
2 downtown_b 30 18 0 0 45 50 60 15 75 68 60 15
3 uptown 25 20 0 5 32 45 55 30 57 65 55 35
4 island 10 30 0 0 40 50 40 0 50 80 40 0
I currently have a dataset with 50,000+ rows of data for which I need to find rolling sums. I have completed this using rollaply which has worked perfectly. I need to apply these rolling sums across a range of widths (600, 1200, 1800...6000) which I have done by cut and pasting each line of script and changing the width. While it works, I'd like to tidy my script but applying a loop, or similar, if possible so that once the rollapply function has completed it's first 'pass' at 600 width, it then completes the same with 1200 and so on. Example:
Var1 Var2 Var3
1 11 19
43 12 1
4 13 47
21 14 29
41 15 42
16 16 5
17 17 16
10 18 15
20 19 41
44 20 27
width_2 <- rollapply(x$Var1, FUN = sum, width = 2)
width_3 <- rollapply(x$Var1, FUN = sum, width = 3)
width_4 <- rollapply(x$Var1, FUN = sum, width = 4)
Is there a way to run widths 2, 3, then 4 in a simpler way rather than cut and paste, particularly when I have up to 10 widths, and then need to run this across other cols. Any help would be appreciated.
We can use lapply in base R
lst1 <- lapply(2:4, function(i) rollapply(x$Var1, FUN = sum, width = i))
names(lst1) <- paste0('width_', 2:4)
list2env(lst1, .GlobalEnv)
NOTE: It is not recommended to create multiple objects in the global environment. Instead, the list would be better
Or with a for loop
for(v in 2:4) {
assign(paste0('width_', v), rollapply(x$Var1, FUN = sum, width = v))
}
Create a function to do this for multiple dataset
f1 <- function(col1, i) {
rollapply(col1, FUN = sum, width = i)
}
lapply(x[c('Var1', 'Var2')], function(x) lapply(2:4, function(i)
f1(x, i)))
Instead of creating separate vectors in global environment probably you can add these as new columns in the already existing dataframe.
Note that rollaplly(..., FUN = sum) is same as rollsum.
library(dplyr)
library(zoo)
bind_cols(x, purrr::map_dfc(2:4,
~x %>% transmute(!!paste0('Var1_roll_', .x) := rollsumr(Var1, .x, fill = NA))))
# Var1 Var2 Var3 Var1_roll_2 Var1_roll_3 Var1_roll_4
#1 1 11 19 NA NA NA
#2 43 12 1 44 NA NA
#3 4 13 47 47 48 NA
#4 21 14 29 25 68 69
#5 41 15 42 62 66 109
#6 16 16 5 57 78 82
#7 17 17 16 33 74 95
#8 10 18 15 27 43 84
#9 20 19 41 30 47 63
#10 44 20 27 64 74 91
You can use seq to generate the variable window size.
seq(600, 6000, 600)
#[1] 600 1200 1800 2400 3000 3600 4200 4800 5400 6000
I want to fill NA rows based on checking the differences between the closest non-NA labeled rows.
For instance
data <- data.frame(sd_value=c(34,33,34,37,36,45),
value=c(383,428,437,455,508,509),
label=c(c("bad",rep(NA,4),"unable")))
> data
sd_value value label
1 34 383 bad
2 33 428 <NA>
3 34 437 <NA>
4 37 455 <NA>
5 36 508 <NA>
6 45 509 unable
I want to evaluate how to change NA rows with checking the difference between sd_value and value those close to bad and unablerows.
if we want to get differences between the rows we can do;
library(dplyr)
data%>%
mutate(diff_val=c(0,diff(value)), diff_sd_val=c(0,diff(sd_value)))
sd_value value label diff_val diff_sd_val
1 34 383 bad 0 0
2 33 428 <NA> 45 -1
3 34 437 <NA> 9 1
4 37 455 <NA> 18 3
5 36 508 <NA> 53 -1
6 45 509 unable 1 9
The condition how I want to label the NA rows is
if the diff_val<50 and diff_sd_val<9 label them with the last non-NA label else use the first non-NA label after the last NA row.
So that the expected output would be
sd_value value label diff_val diff_sd_val
1 34 383 bad 0 0
2 33 428 bad 45 -1
3 34 437 bad 9 1
4 37 455 bad 18 3
5 36 508 unable 53 -1
6 45 509 unable 1 9
The possible solution I cooked up so far:
custom_labelling <- function(x,y,label){
diff_sd_val<-c(NA,diff(x))
diff_val<-c(NA,diff(y))
label <- NA
for (i in 1:length(label)){
if(is.na(label[i])&diff_sd_val<9&diff_val<50){
label[i] <- label
}
else {
label <- label[i]
}
}
return(label)
}
which gives
data%>%
mutate(diff_val=c(0,diff(value)), diff_sd_val=c(0,diff(sd_value)))%>%
mutate(custom_label=custom_labelling(sd_value,value,label))
Error in mutate_impl(.data, dots) :
Evaluation error: missing value where TRUE/FALSE needed.
In addition: Warning message:
In if (is.na(label[i]) & diff_sd_val < 9 & diff_val < 50) { :
the condition has length > 1 and only the first element will be used
One option is to find NA and non-NA index and based on the condition select the closest label to it.
library(dplyr)
#Create a new dataframe with diff_val and diff_sd_val
data1 <- data%>% mutate(diff_val=c(0,diff(value)), diff_sd_val=c(0,diff(sd_value)))
#Get the NA indices
NA_inds <- which(is.na(data1$label))
#Get the non-NA indices
non_NA_inds <- setdiff(1:nrow(data1), NA_inds)
#For every NA index
for (i in NA_inds) {
#Check the condition
if(data1$diff_sd_val[i] < 9 & data1$diff_val[i] < 50)
#Get the last non-NA label
data1$label[i] <- data1$label[non_NA_inds[which.max(i > non_NA_inds)]]
else
#Get the first non-NA label after last NA value
data1$label[i] <- data1$label[non_NA_inds[i < non_NA_inds]]
}
data1
# sd_value value label diff_val diff_sd_val
#1 34 383 bad 0 0
#2 33 428 bad 45 -1
#3 34 437 bad 9 1
#4 37 455 bad 18 3
#5 36 508 unable 53 -1
#6 45 509 unable 1 9
You can remove diff_val and diff_sd_val columns later if not needed.
We can also create a function
custom_label <- function(label, diff_val, diff_sd_val) {
NA_inds <- which(is.na(label))
non_NA_inds <- setdiff(1:length(label), NA_inds)
new_label = label
for (i in NA_inds) {
if(diff_sd_val[i] < 9 & diff_val[i] < 50)
new_label[i] <- label[non_NA_inds[which.max(i > non_NA_inds)]]
else
new_label[i] <- label[non_NA_inds[i < non_NA_inds]]
}
return(new_label)
}
and then apply it
data%>%
mutate(diff_val = c(0, diff(value)),
diff_sd_val = c(0, diff(sd_value)),
new_label = custom_label(label, diff_val, diff_sd_val))
# sd_value value label diff_val diff_sd_val new_label
#1 34 383 bad 0 0 bad
#2 33 428 <NA> 45 -1 bad
#3 34 437 <NA> 9 1 bad
#4 37 455 <NA> 18 3 bad
#5 36 508 <NA> 53 -1 unable
#6 45 509 unable 1 9 unable
If we want to apply it by group we can add a group_by statement and it should work.
data%>%
group_by(group) %>%
mutate(diff_val = c(0, diff(value)),
diff_sd_val = c(0, diff(sd_value)),
new_label = custom_label(label, diff_val, diff_sd_val))
I got the following dataframe called nodes_df:
x y node_demand
1 2 62 3
2 80 25 14
3 36 88 1
4 57 23 14
5 33 17 19
6 76 43 2
7 77 85 14
8 94 6 6
10 59 72 6
. . . .
. . . .
. . . .
. . . .
45 60 84 8
46 35 100 5
47 38 2 1
48 9 9 7
50 1 58 2
I have to split this dataframe between hubs and clients.
hubs <- nodes_df[keep <- sample(1:total_nodes, requested_hubs, replace = FALSE),]
client_nodes <- nodes_df[-keep, ]
I need to randomly select 1 row at a time from clients_nodes and calculate the total node_demand, I need to keep adding rows until random_clients$node_demand exceedes 120.
random_clients <- client_nodes[sample(nrow(client_nodes), size = 1, replace = FALSE),]
I created the following variables and while loop
node_demand <- c(0)
cumulative_demand <- cumsum(node_demand)
client_nodes <- nodes_df[-keep, ]
last_node <- cumsum(cumulative_demand) >= max_supply_capacity
condition = TRUE
while(condition){
random_clients <- client_nodes[sample(nrow(client_nodes), size = 1, replace = FALSE),]
node_demand <- c(node_demand,random_clients$node_demand)
cumulative_demand <- cumsum(node_demand)
if(cumulative_demand <= max_supply_capacity){
condition == FALSE
}
}
The loop doesn't stop and I get the following return value:
cumulative_demand
[1] 0 14 20 26 27 35 49 50 68 79 97 100 101 104 109 118
[17] 119 137 150 164 178 185 188 191 208 209 219 222 227 246 252 272 (it carries on and on)
I am not sure why the loop doesn't stop despite the condition cumulative_demand <= max_supply_capacity being met.
Anybody could show me how to fix it?
I managed to fix it :).
I had to use ifelse() so R could evaluate the condition of a vector. The normal if() statement wouldn't work in this case
while(TRUE){
random_clients <- client_nodes[sample(nrow(client_nodes), size = 1, replace = FALSE),]
node_demand <- c(node_demand,random_clients$node_demand)
cumulative_demand <- cumsum(node_demand)
last_node <- (cumulative_demand <= max_supply_capacity)
ifelse(last_node == FALSE,break,next)
}
I had to use and ifelse() instead of an if() statement as shown in the problem description.
I have a data.frame
set.seed(100)
exp <- data.frame(exp = c(rep(LETTERS[1:2], each = 10)), re = c(rep(seq(1, 10, 1), 2)), age1 = seq(10, 29, 1), age2 = seq(30, 49, 1),
h = c(runif(20, 10, 40)), h2 = c(40 + runif(20, 4, 9)))
I'd like to make a lm for each row in a data set (h and h2 ~ age1 and age2)
I do it by loop
exp$modelh <- 0
for (i in 1:length(exp$exp)){
age = c(exp$age1[i], exp$age2[i])
h = c(exp$h[i], exp$h2[i])
model = lm(age ~ h)
exp$modelh[i] = coef(model)[1] + 100 * coef(model)[2]
}
and it works well but takes some time with very large files. Will be grateful for the faster solution f.ex. dplyr
Using dplyr, we can try with rowwise() and do. Inside the do, we concatenate (c) the 'age1', 'age2' to create 'age', likewise, we can create 'h', apply lm, extract the coef to create the column 'modelh'.
library(dplyr)
exp %>%
rowwise() %>%
do({
age <- c(.$age1, .$age2)
h <- c(.$h, .$h2)
model <- lm(age ~ h)
data.frame(., modelh = coef(model)[1] + 100*coef(model)[2])
} )
gives the output
# exp re age1 age2 h h2 modelh
#1 A 1 10 30 19.23298 46.67906 68.85506
#2 A 2 11 31 17.73018 47.55402 66.17050
#3 A 3 12 32 26.56967 46.69174 84.98486
#4 A 4 13 33 11.69149 47.74486 61.98766
#5 A 5 14 34 24.05648 46.10051 82.90167
#6 A 6 15 35 24.51312 44.85710 89.21053
#7 A 7 16 36 34.37208 47.85151 113.37492
#8 A 8 17 37 21.10962 48.40977 74.79483
#9 A 9 18 38 26.39676 46.74548 90.34187
#10 A 10 19 39 15.10786 45.38862 75.07002
#11 B 1 20 40 28.74989 46.44153 100.54666
#12 B 2 21 41 36.46497 48.64253 125.34773
#13 B 3 22 42 18.41062 45.74346 81.70062
#14 B 4 23 43 21.95464 48.77079 81.20773
#15 B 5 24 44 32.87653 47.47637 115.95097
#16 B 6 25 45 30.07065 48.44727 101.10688
#17 B 7 26 46 16.13836 44.90204 84.31080
#18 B 8 27 47 20.72575 47.14695 87.00805
#19 B 9 28 48 20.78425 48.94782 84.25406
#20 B 10 29 49 30.70872 44.65144 128.39415
We could do this with the devel version of data.table i.e. v1.9.5. Instructions to install the devel version are here.
We convert the 'data.frame' to 'data.table' (setDT), create a column 'rn' with the option keep.rownames=TRUE. We melt the dataset by specifying the patterns in the measure to convert from 'wide' to 'long' format. Grouped by 'rn', we do the lm and get the coef. This can be assigned as a new column in the original dataset ('exp') while removing the unwanted 'rn' column by assigning (:=) it to NULL.
library(data.table)#v1.9.5+
modelh <- melt(setDT(exp, keep.rownames=TRUE), measure=patterns('^age', '^h'),
value.name=c('age', 'h'))[, {model <- lm(age ~h)
coef(model)[1] + 100 * coef(model)[2]},rn]$V1
exp[, modelh:= modelh][, rn := NULL]
exp
# exp re age1 age2 h h2 modelh
# 1: A 1 10 30 19.23298 46.67906 68.85506
# 2: A 2 11 31 17.73018 47.55402 66.17050
# 3: A 3 12 32 26.56967 46.69174 84.98486
# 4: A 4 13 33 11.69149 47.74486 61.98766
# 5: A 5 14 34 24.05648 46.10051 82.90167
# 6: A 6 15 35 24.51312 44.85710 89.21053
# 7: A 7 16 36 34.37208 47.85151 113.37492
# 8: A 8 17 37 21.10962 48.40977 74.79483
# 9: A 9 18 38 26.39676 46.74548 90.34187
#10: A 10 19 39 15.10786 45.38862 75.07002
#11: B 1 20 40 28.74989 46.44153 100.54666
#12: B 2 21 41 36.46497 48.64253 125.34773
#13: B 3 22 42 18.41062 45.74346 81.70062
#14: B 4 23 43 21.95464 48.77079 81.20773
#15: B 5 24 44 32.87653 47.47637 115.95097
#16: B 6 25 45 30.07065 48.44727 101.10688
#17: B 7 26 46 16.13836 44.90204 84.31080
#18: B 8 27 47 20.72575 47.14695 87.00805
#19: B 9 28 48 20.78425 48.94782 84.25406
#20: B 10 29 49 30.70872 44.65144 128.39415
Great (double) answer from #akrun.
Just a suggestion for your future analysis as you mentioned "it's an example of a bigger problem". Obviously, if you are really interested in building models rowwise then you'll create more and more columns as your age and h observations increase. If you get N observations you'll have to use 2xN columns for those 2 variables only.
I'd suggest to use a long data format in order to increase your rows instead of your columns.
Something like:
exp[1,] # how your first row (model building info) looks like
# exp re age1 age2 h h2
# 1 A 1 10 30 19.23298 46.67906
reshape(exp[1,], # how your model building info is transformed
varying = list(c("age1","age2"),
c("h","h2")),
v.names = c("age_value","h_value"),
direction = "long")
# exp re time age_value h_value id
# 1.1 A 1 1 10 19.23298 1
# 1.2 A 1 2 30 46.67906 1
Apologies if the "bigger problem" refers to something else and this answer is irrelevant.
With base R, the function sprintf can help us create formulas. And lapply carries out the calculation.
strings <- sprintf("c(%f,%f) ~ c(%f,%f)", exp$age1, exp$age2, exp$h, exp$h2)
lst <- lapply(strings, function(x) {model <- lm(as.formula(x));coef(model)[1] + 100 * coef(model)[2]})
exp$modelh <- unlist(lst)
exp
# exp re age1 age2 h h2 modelh
# 1 A 1 10 30 19.23298 46.67906 68.85506
# 2 A 2 11 31 17.73018 47.55402 66.17050
# 3 A 3 12 32 26.56967 46.69174 84.98486
# 4 A 4 13 33 11.69149 47.74486 61.98766
# 5 A 5 14 34 24.05648 46.10051 82.90167
# 6 A 6 15 35 24.51312 44.85710 89.21053
# 7 A 7 16 36 34.37208 47.85151 113.37493
# 8 A 8 17 37 21.10962 48.40977 74.79483
# 9 A 9 18 38 26.39676 46.74548 90.34187
# 10 A 10 19 39 15.10786 45.38862 75.07002
# 11 B 1 20 40 28.74989 46.44153 100.54666
# 12 B 2 21 41 36.46497 48.64253 125.34773
# 13 B 3 22 42 18.41062 45.74346 81.70062
# 14 B 4 23 43 21.95464 48.77079 81.20773
# 15 B 5 24 44 32.87653 47.47637 115.95097
# 16 B 6 25 45 30.07065 48.44727 101.10688
# 17 B 7 26 46 16.13836 44.90204 84.31080
# 18 B 8 27 47 20.72575 47.14695 87.00805
# 19 B 9 28 48 20.78425 48.94782 84.25406
# 20 B 10 29 49 30.70872 44.65144 128.39416
In the lapply function the expression as.formula(x) is what converts the formulas created in the first line into a format usable by the lm function.
Benchmark
library(dplyr)
library(microbenchmark)
set.seed(100)
big.exp <- data.frame(age1=sample(30, 1e4, T),
age2=sample(30:50, 1e4, T),
h=runif(1e4, 10, 40),
h2= 40 + runif(1e4,4,9))
microbenchmark(
plafort = {strings <- sprintf("c(%f,%f) ~ c(%f,%f)", big.exp$age1, big.exp$age2, big.exp$h, big.exp$h2)
lst <- lapply(strings, function(x) {model <- lm(as.formula(x));coef(model)[1] + 100 * coef(model)[2]})
big.exp$modelh <- unlist(lst)},
akdplyr = {big.exp %>%
rowwise() %>%
do({
age <- c(.$age1, .$age2)
h <- c(.$h, .$h2)
model <- lm(age ~ h)
data.frame(., modelh = coef(model)[1] + 100*coef(model)[2])
} )}
,times=5)
t: seconds
expr min lq mean median uq max neval cld
plafort 13.00605 13.41113 13.92165 13.56927 14.53814 15.08366 5 a
akdplyr 26.95064 27.64240 29.40892 27.86258 31.02955 33.55940 5 b
(Note: I downloaded the newest 1.9.5 devel version of data.table today, but continued to receive errors when trying to test it.
The results also differ fractionally (1.93 x 10^-8). Rounding likely accounts for the difference.)
all.equal(pl, ak)
[1] "Attributes: < Component “class”: Lengths (1, 3) differ (string compare on first 1) >"
[2] "Attributes: < Component “class”: 1 string mismatch >"
[3] "Component “modelh”: Mean relative difference: 1.933893e-08"
Conclusion
The lapply approach seems to perform well compared to dplyr with respect to speed, but it's 5 digit rounding may be an issue. Improvements may be possible. Perhaps using apply after converting to matrix to increase speed and efficiency.