I'm trying to add a column which reads my dataframe's column and outputs a 1 if the element is bigger than a certain number (and a zero if the condition isn't met). However, this code doesn't seem to work: df is an existing dataframe.
df2 <- data.frame(df2, C=Recode(df$numbers, "hi:200=1; else=0")) ##C = numbers > 200 = 1
I'm using R's car library.
Does this achieve what you need?
df2 <- tibble(numbers = c(1, 200, 201))
df2$recoded <- ifelse(df2$numbers > 200, 1, 0)
df2
# # A tibble: 3 x 2
# numbers recoded
# <dbl> <dbl>
# 1 1 0
# 2 200 0
# 3 201 1
In base R we can also do
df2$recoded <- as.integer(df2$numbers > 200)
In data.table we could do:
library(data.table)
df <- datasets::cars
setDT(df)
df[, numbers := ifelse(df$dist > 10, 1, 0)][1:10, ]
#> speed dist numbers
#> 1: 4 2 0
#> 2: 4 10 0
#> 3: 7 4 0
#> 4: 7 22 1
#> 5: 8 16 1
#> 6: 9 10 0
#> 7: 10 18 1
#> 8: 10 26 1
#> 9: 10 34 1
#> 10: 11 17 1
Created on 2021-03-17 by the reprex package (v0.3.0)
Related
I have a table with only one column and more than 200 rows. It includes three values, 0, 1 and 3. I´m interested in only these incidents, where an 1 follwos a 0. Can R count all X=1 if X-1 = =, given that X is the value of any row.
It would be great, if someone could help !
Best, Anna
Do you mean something like this?
# Create some sample data
set.seed(2020)
df <- data.frame(incident = sample(c(0, 1, 3), 10, replace = TRUE))
# incident
#1 3
#2 1
#3 0
#4 0
#5 1
#6 1
#7 0
#8 0
#9 1
#10 1
sum(c(df$incident[-1] == 1, FALSE) * (df$incident == 0))
# Or: with(df, sum(c(incident[-1] == 1, FALSE) * (incident == 0)))
#[1] 2
Here, c(incident[-1] == 1, FALSE) * (incident == 0) is the logical AND of x[i-1] = 0 and x[i] = 1. sum then sums the number of occurrences (in this case there are 2: one in rows 4/5 and one in rows 8/9).
library(tidyverse)
set.seed(123)
(df <- tibble(value = sample(c(0, 1, 3),size = 200, replace = TRUE)))
#> # A tibble: 200 x 1
#> value
#> <dbl>
#> 1 3
#> 2 3
#> 3 3
#> 4 1
#> 5 3
#> 6 1
#> 7 1
#> 8 1
#> 9 3
#> 10 0
#> # … with 190 more rows
count <- 0
#use map instead of walk to view the process row by row
walk(2:nrow(df), ~ {
if (df$value[[.x - 1]] == 0 && df$value[[.x]] == 1) count <<- count + 1
})
count
#> [1] 26
#some rows where the pattern is happening
df[86:87, ]
#> # A tibble: 2 x 1
#> value
#> <dbl>
#> 1 0
#> 2 1
df[93:94, ]
#> # A tibble: 2 x 1
#> value
#> <dbl>
#> 1 0
#> 2 1
Created on 2021-06-28 by the reprex package (v2.0.0)
Using dplyr:
transmute(df, dif = c(NA, diff(value))) %>%
count(dif) %>%
filter(dif == 1)
#> # A tibble: 1 x 2
#> dif n
#> <dbl> <int>
#> 1 1 26
Created on 2021-06-28 by the reprex package (v2.0.0)
I have a tibble dt given as follows:
library(tidyverse)
dt <- tibble(x=as.integer(c(0,0,1,0,0,0,1,1,0,1))) %>%
mutate(grp = as.factor(c(rep("A",3), rep("B",4), rep("C",1), rep("D",2))))
dt
As one can observe the rule for grouping is:
starts 0 and ends with 1 (e.g., groups A, B, D) or
it solely contains 1 (e.g., group C)
Problem: Given a tibble with column integer vector x of zeros and 1 that starts with 0 and ends in 1, what is the most efficient way to obtain a grouping using R? (You can use any grouping symbols/factors.)
We can get the cumulative sum of 'x' (assuming it is binary), take the lag add 1 and use that index to replace it with LETTERS (Note that LETTERS was used only as part of matching with the expected output - it can take go up to certain limit)
library(dplyr)
dt %>%
mutate(grp2 = LETTERS[lag(cumsum(x), default = 0)+ 1])
-output
# A tibble: 10 x 3
x grp grp2
<int> <fct> <chr>
1 0 A A
2 0 A A
3 1 A A
4 0 B B
5 0 B B
6 0 B B
7 1 B B
8 1 C C
9 0 D D
10 1 D D
Though the strategy proposed by Akrun is fantastic, yet to show that it can be managed through accumulate also
library(tidyverse)
dt <- tibble(x=as.integer(c(0,0,1,0,0,0,1,1,0,1))) %>%
mutate(grp = as.factor(c(rep("A",3), rep("B",4), rep("C",1), rep("D",2))))
dt %>%
mutate(GRP = accumulate(lag(x, default = 0),.init =1, ~ if(.y != 1) .x else .x+1)[-1])
#> # A tibble: 10 x 3
#> x grp GRP
#> <int> <fct> <dbl>
#> 1 0 A 1
#> 2 0 A 1
#> 3 1 A 1
#> 4 0 B 2
#> 5 0 B 2
#> 6 0 B 2
#> 7 1 B 2
#> 8 1 C 3
#> 9 0 D 4
#> 10 1 D 4
Created on 2021-06-13 by the reprex package (v2.0.0)
I am looking for a concise way to filter a data.frame for all rows smaller than a value x with all following values also smaller than x. I found a way but it is somehwat verbose. I tried to do it with dplyr::cumall and cumany, but was not able to figure it out.
Here is a small reprex including my actual approach. Ideally I would only have one filter line or mutate + filter, but with the current approach it takes two rounds of mutate/filter.
library(dplyr)
# Original data
tbl <- tibble(value = c(100,100,100,10,10,5,10,10,5,5,5,1,1,1,1))
# desired output:
# keep only rows, where value is smaller than 5 and ...
# no value after that is larger than 5
tbl %>%
mutate(id = row_number()) %>%
filter(value <= 5) %>%
mutate(id2 = lead(id, default = max(id) + 1) - id) %>%
filter(id2 == 1)
#> # A tibble: 7 x 3
#> value id id2
#> <dbl> <int> <dbl>
#> 1 5 9 1
#> 2 5 10 1
#> 3 5 11 1
#> 4 1 12 1
#> 5 1 13 1
#> 6 1 14 1
#> 7 1 15 1
Created on 2020-04-20 by the reprex package (v0.3.0)
You could combine cummin with a reversed reverse cummax:
tbl %>% filter(rev(cummax(rev(value))) <= 5 & cummin(value) <= 5)
# A tibble: 7 x 1
value
<dbl>
1 5
2 5
3 5
4 1
5 1
6 1
7 1
A base R option is to use subset + rle
tblout <- subset(tbl,
with(rle(value<=5 & c(0,diff(value))<=0),
rep(lengths>1 & values,lengths)))
such that
> tblout
# A tibble: 7 x 1
value
<dbl>
1 5
2 5
3 5
4 1
5 1
6 1
7 1
I have a data.frame of (sub)string positions within a larger string. The data contains the start of a (sub)string and it's length. The end position of the (sub)string can be easily calculated.
data1 <- data.frame(start = c(1,3,4,9,10,13),
length = c(2,1,3,1,2,1)
)
data1$end <- (data1$start + data1$length - 1)
data1
#> start length end
#> 1 1 2 2
#> 2 3 1 3
#> 3 4 3 6
#> 4 9 1 9
#> 5 10 2 11
#> 6 13 1 13
Created on 2019-12-10 by the reprex package (v0.3.0)
I would like to 'compress' this data.frame by summarizing continuous (sub)strings (strings that are connected with each other) so that my new data looks like this:
data2 <- data.frame(start = c(1,9,13),
length = c(6,3,1)
)
data2$end <- (data2$start + data2$length - 1)
data2
#> start length end
#> 1 1 6 6
#> 2 9 3 11
#> 3 13 1 13
Created on 2019-12-10 by the reprex package (v0.3.0)
Is there preferably a base R solution which gets me from data1 to data2?
f = cumsum(with(data1, c(0, start[-1] - head(end, -1))) != 1)
do.call(rbind, lapply(split(data1, f), function(x){
with(x, data.frame(start = start[1],
length = tail(end, 1) - start[1] + 1,
end = tail(end, 1)))}))
# start length end
#1 1 6 6
#2 9 3 11
#3 13 1 13
Using dplyr we can do the following:
library(dplyr)
data1 %>%
group_by(consecutive = cumsum(start != lag(end, default = 0) + 1)) %>%
summarise(start = min(start), length=sum(length), end=max(end)) %>%
ungroup %>% select(-consecutive)
#> # A tibble: 3 x 3
#> start length end
#> <dbl> <dbl> <dbl>
#> 1 1 6 6
#> 2 9 3 11
#> 3 13 1 13
let's say I have a data frame which looks something like this
A <- c(1:100)
B <- c(0.5:100)
df <- data.frame(A,B)
And I want to get 25 random rows out of this data frame with
df[sample(nrow(df), size = 25, replace = FALSE),]
But now I want to repeat this sample function 100 times and save every result individually.
I've tried to use the repeat function but I can't find a way to save every result.
Thank you.
As mentioned in the comments, the replicate implementation can reach your goal, i.e.,
res <- replicate(100,df[sample(nrow(df), size = 25, replace = FALSE),],simplify = F)
An alternative is to use sapply (or lapply), i.e.,
res <- sapply(1:100, function(k) df[sample(nrow(df), size = 25, replace = FALSE),],simplify = F)
or
res <- lapply(1:100, function(k) df[sample(nrow(df), size = 25, replace = FALSE),])
replicate() is a great option for this problem.
If you would like your final results in a single table with a column for the ID variable, you can use bind_rows() from the dplyr package. Here is a smaller example (3 samples from a data set of 10 rows) that may allow easier understanding of replicate()'s behavior:
library(dplyr, warn.conflicts = FALSE)
# make a smaller data set of 10 rows
d <- data.frame(
A = 1:10,
B = LETTERS[1:10]
) %>% print
#> A B
#> 1 1 A
#> 2 2 B
#> 3 3 C
#> 4 4 D
#> 5 5 E
#> 6 6 F
#> 7 7 G
#> 8 8 H
#> 9 9 I
#> 10 10 J
# create 3 samples, with each sample containing 4 rows
reps <- replicate(3, d[sample(nrow(d), 4, FALSE), ], simplify = FALSE) %>% print
#> [[1]]
#> A B
#> 2 2 B
#> 5 5 E
#> 6 6 F
#> 1 1 A
#>
#> [[2]]
#> A B
#> 3 3 C
#> 2 2 B
#> 5 5 E
#> 8 8 H
#>
#> [[3]]
#> A B
#> 4 4 D
#> 9 9 I
#> 3 3 C
#> 8 8 H
# bind the list elements into a single tibble, with an ID column for the sample
bind_rows(reps, .id = "sample_id")
#> sample_id A B
#> 1 1 2 B
#> 2 1 5 E
#> 3 1 6 F
#> 4 1 1 A
#> 5 2 3 C
#> 6 2 2 B
#> 7 2 5 E
#> 8 2 8 H
#> 9 3 4 D
#> 10 3 9 I
#> 11 3 3 C
#> 12 3 8 H
Created on 2019-12-02 by the reprex package (v0.3.0)