I need to check whether the number of elements of each unique value in the variable PPT in A is equal to the number of elements of each unique value in PPT in B, and whether there is any value unique only to A or only to B.
For example:
PPTa <- c("ppt0100109","ppt0301104","ppt0100109","ppt0100109","ppt0300249","ppt0100109","ppt0300249","ppt0100109","ppt0504409","ppt2303401","ppt0704210","ppt0704210","ppt0100109")
CNa <- c(110,54,110,110,49,10,49,110,409,40,10,10,110)
LLa <- c(150,55,150,150,45,15,45,115,405,45,5,15,50)
A <-data.frame(PPTa,CNa,LLa)
PPTb <- c("ppt0100200","ppt0300249","ppt0100109","ppt0300249","ppt0100109","ppt0764091","ppt2303401","ppt0704210","ppt0704210","ppt0100109")
CNb <- c(110,54,110,110,49,10,49,110,409,40)
LLb <- c(150,55,150,150,45,15,45,115,405,45)
B <-data.frame(PPTb,CNb,LLb)
In this case, we have these unique values which occur a certain amount of times:
A$PPTa TIMES
"ppt0100109" 6
"ppt0301104" 1
"ppt0300249" 2
"ppt0504409" 1
"ppt2303401" 1
"ppt0704210" 2
B$PPTb TIMES
"ppt0100200" 1
"ppt0300249" 2
"ppt0100109" 3
"ppt0764091" 1
"ppt2303401" 1
"ppt0704210" 2
I would like to create a new matrix (or anything you could suggest) with a value of 0 if the unique value exists both in A and B with the same number of elements, a value of 1 if it exists in both dataframes A and B but the number of elements differ, and a value of 2 if the value exists only in one of the two dataframes.
Something like:
A$PPTa TIMES OUTPUT
"ppt0100109" 6 1
"ppt0301104" 1 2
"ppt0300249" 2 0
"ppt0504409" 1 2
"ppt2303401" 1 0
"ppt0704210" 2 0
B$PPTb TIMES OUTPUT
"ppt0100200" 1 2
"ppt0300249" 2 0
"ppt0100109" 3 1
"ppt0764091" 1 2
"ppt2303401" 1 0
"ppt0704210" 2 0
You can use a nested ifelse statement,
ifelse(do.call(paste0, A) %in% do.call(paste0, B), 0, ifelse(A$PPTa %in% B$PPTb, 1, 2))
#[1] 1 0 2 2 0 0
ifelse(do.call(paste0, B) %in% do.call(paste0, A), 0, ifelse(B$PPTb %in% A$PPTa, 1, 2))
#[1] 1 2 0 0 2 0
Related
I am trying to combine 2 or more different columns, with the replacement of values. For example,
a b
1 0
1 1
<NA> 1
0 1
0 0
would become
c
1
1
1
1
0
Most functions seem to have an index column, which would be different from having and overwriting value.
Is there a possible way to combine with replacement according to value?
We could do it with an ifelse statement:
library(dplyr)
df %>%
#mutate(c = ifelse(is.na(a) | a ==1, 1, b))
transmute(c = ifelse(is.na(a) | a ==1, 1, b))
c
1 1
2 1
3 1
4 1
5 0
I have a vector of numbers in a data.frame such as below.
df <- data.frame(a = c(1,2,3,4,2,3,4,5,8,9,10,1,2,1))
I need to create a new column which gives a running count of entries that are greater than their predecessor. The resulting column vector should be this:
0,1,2,3,0,1,2,3,4,5,6,0,1,0
My attempt is to create a "flag" column of diffs to mark when the values are greater.
df$flag <- c(0,diff(df$a)>0)
> df$flag
0 1 1 1 0 1 1 1 1 1 1 0 1 0
Then I can apply some dplyr group/sum magic to almost get the right answer, except that the sum doesn't reset when flag == 0:
df %>% group_by(flag) %>% mutate(run=cumsum(flag))
a flag run
1 1 0 0
2 2 1 1
3 3 1 2
4 4 1 3
5 2 0 0
6 3 1 4
7 4 1 5
8 5 1 6
9 8 1 7
10 9 1 8
11 10 1 9
12 1 0 0
13 2 1 10
14 1 0 0
I don't want to have to resort to a for() loop because I have several of these running sums to compute with several hundred thousand rows in a data.frame.
Here's one way with ave:
ave(df$a, cumsum(c(F, diff(df$a) < 0)), FUN=seq_along) - 1
[1] 0 1 2 3 0 1 2 3 4 5 6 0 1 0
We can get a running count grouped by diff(df$a) < 0. Which are the positions in the vector that are less than their predecessors. We add c(F, ..) to account for the first position. The cumulative sum of that vector creates an index for grouping. The function ave can carry out a function on that index, we use seq_along for a running count. But since it starts at 1, we subtract by one ave(...) - 1 to start from zero.
A similar approach using dplyr:
library(dplyr)
df %>%
group_by(cumsum(c(FALSE, diff(a) < 0))) %>%
mutate(row_number() - 1)
You don't need dplyr:
fun <- function(x) {
test <- diff(x) > 0
y <- cumsum(test)
c(0, y - cummax(y * !test))
}
fun(df$a)
[1] 0 1 2 3 0 1 2 3 4 5 6 0 1 0
a <- c(1,2,3,4,2,3,4,5,8,9,10,1,2,1)
f <- c(0, diff(a)>0)
ifelse(f, cumsum(f), f)
that it is without reset.
with reset:
unlist(tapply(f, cumsum(c(0, diff(a) < 0)), cumsum))
I have a dataframe, which contains 100.000 rows. It looks like this:
Value
1
2
-1
-2
0
3
4
-1
3
I want to create an extra column (column B). Which consist of 0 and 1's.
It is basically 0, but when there are 5 data points in a row positive OR negative, then it should give a 1. But, only if they are in a row (e.g.: when the row is positive, and there is a negative number.. the count shall start again).
Value B
1 0
2 0
1 0
2 0
2 1
3 1
4 1
-1 0
3 0
I tried different loops, but It didn't work. I also tried to convert the whole DF to a list (and loop over the list). Unfortunately with no end.
Here's an approach that uses the rollmean function from the zoo package.
set.seed(1000)
df = data.frame(Value = sample(-9:9,1000,replace=T))
sign = sign(df$Value)
library(zoo)
rolling = rollmean(sign,k=5,fill=0,align="right")
df$B = as.numeric(abs(rolling) == 1)
I generated 1000 values with positive and negative sets.
Extract the sign of the values - this will be -1 for negative, 1 for positive and 0 for 0
Calculate the right aligned rolling mean of 5 values (it will average x[1:5], x[2:6], ...). This will be 1 or -1 if all the values in a row are positive or negative (respectively)
Take the absolute value and store the comparison against 1. This is a logical vector that turns into 0s and 1s based on your conditions.
Note - there's no need for loops. This can all be vectorised (once we have the rolling mean calculated).
This will work. Not the most efficient way to do it but the logic is pretty transparent -- just check if there's only one unique sign (i.e. +, -, or 0) for each sequence of five adjacent rows:
dat <- data.frame(Value=c(1,2,1,2,2,3,4,-1,3))
dat$new_col <- NA
dat$new_col[1:4] <- 0
for (x in 5:nrow(dat)){
if (length(unique(sign(dat$Value[(x-4):x])))==1){
dat$new_col[x] <- 1
} else {
dat$new_col[x] <- 0
}
}
Use the cumsum(...diff(...) <condition>) idiom to create a grouping variable, and ave to calculate the indices within each group.
d$B2 <- ave(d$Value, cumsum(c(0, diff(sign(d$Value)) != 0)), FUN = function(x){
as.integer(seq_along(x) > 4)})
# Value B B2
# 1 1 0 0
# 2 2 0 0
# 3 1 0 0
# 4 2 0 0
# 5 2 1 1
# 6 3 1 1
# 7 4 1 1
# 8 -1 0 0
# 9 3 0 0
I want a simple way to create a new variable determining whether a boolean is ever true in R data frame.
Here is and example:
Suppose in the dataset I have 2 variables (among other variables which are not relevant) 'a' and 'b' and 'a' determines a group, while 'b' is a boolean with values TRUE (1) or FALSE (0). I want to create a variable 'c', which is also a boolean being 1 for all entries in groups where 'b' is at least once 'TRUE', and 0 for all entries in groups in which 'b' is never TRUE.
From entries like below:
a b
-----
1 1
2 0
1 0
1 0
1 1
2 0
2 0
3 0
3 1
3 0
I want to get variable 'c' like below:
a b c
-----------
1 1 1
2 0 0
1 0 1
1 0 1
1 1 1
2 0 0
2 0 0
3 0 1
3 1 1
3 0 1
-----------
I know how to do it in Stata, but I haven't done similar things in R yet, and it is difficult to find information on that on the internet.
In fact I am doing that only in order to later remove all the observations for which 'c' is 0, so any other suggestions would be fine as well. The application of that relates to multinomial logit estimation, where the alternatives that are never-chosen need to be removed from the dataset before estimation.
if X is your data frame
library(dplyr)
X <- X %>%
group_by(a) %>%
mutate(c = any(b == 1))
A base R option would be
df1$c <- with(df1, ave(b, a, FUN=any))
Or
library(sqldf)
sqldf('select * from df1
left join(select a, b,
(sum(b))>0 as c
from df1
group by a)
using(a)')
Simple data.table approach
require(data.table)
data <- data.table(data)
data[, c := any(b), by = a]
Even though logical and numeric (0-1) columns behave identically for all intents and purposes, if you'd like a numeric result you can simply wrap the call to any with as.numeric.
An answer with base R, assuming a and b are in dataframe x
c value is a 1-to-1 mapping with a, and I create a mapping here
cmap <- ifelse(sapply(split(x, x$a), function(x) sum(x[, "b"])) > 0, 1, 0)
Then just add in the mapped value into the data frame
x$c <- cmap[x$a]
Final output
> x
a b c
1 1 1 1
2 2 0 0
3 1 0 1
4 1 0 1
5 1 1 1
6 2 0 0
7 2 0 0
8 3 0 1
9 3 1 1
10 3 0 1
edited to change call to split.
I have a problem very similar to that described here:
subset of data.frame columns to maximize "complete" observations
I am trying to schedule a workshop that will meet five times. I have ten days from which to choose meeting dates, each day having three overlapping possible meeting times. Hence, I have 30 columns grouped into ten groups (days) of three columns (meeting times) each. I need to select 5 columns (or meeting date–time combinations) subject to the following criteria: only one meeting time is selected per day (one column per group); the number of respondents (rows) who can attend all 5 meetings is maximized. Ideally, I would also want to know how the optimal column selection changes if I relax the criterion that respondents must attend ALL 5 meetings, requiring only that they attend 4, or 3, etc.
For simple visualization, let's say I want to know which two columns I should choose—no more than one each from V1, V2, and V3—such that I maximize the number of rows that have no zeros (i.e. row sums to 2).
V1A V1B V1C V2A V2B V2C V3A V3B V3C
1 0 1 0 1 1 1 0 1
1 1 0 0 1 1 0 1 1
0 0 1 1 1 0 0 1 1
1 1 1 1 0 0 1 0 0
1 0 0 0 1 1 0 1 0
0 1 1 0 1 1 0 0 0
1 0 1 1 1 0 1 0 1
The actual data are here: https://drive.google.com/file/d/0B03dE9-8088aMklOUVhuV3gtRHc/view Groups are mon1* tue1* [...] mon2* tue2* [...] fri2*.
The code proposed in the link above would solve my problem if it were not the case that I needed to select columns from groups. Ideally, I would also be able to say which columns I should choose to maximize the number of rows under the weaker condition that a row could have one zero (i.e. row sums to 5 or 4 or 3, etc.).
Many thanks!
You could use rowSums to get the index of rows that have greater than or equal to two 1's. (The conditions are not very clear)
lapply(split(names(df),sub('.$', '', names(df))),
function(x) which(rowSums(df[x])>=2))
#$V1
#[1] 1 2 4 6 7
#$V2
#[1] 1 2 3 5 6 7
#$V3
#[1] 1 2 3 7
This just finds the first column index with 1 (or very first if all zero) in each of three groups, returning a three column matrix, one column for each group.
f <- substring(colnames(df), 1L, nchar(colnames(df))-1L)
ans <- lapply(split(as.list(df), f),
function(x) max.col(do.call(cbind, x), ties.method="first"))
do.call(cbind, ans)
With your dataset this delivers the rows that satisfy the requirement to deliver all rows==1:
> lapply( 1:3, function(grp) which( apply( dat[, grep(grp, names(dat))] , 1,
function(z) sum(z, na.rm=TRUE)==3) ) )
[[1]]
[1] 4
[[2]]
integer(0)
[[3]]
integer(0)
If you relax the requirement to allow values less than 3 you get more candidates:
> lapply( 1:3, function(grp) which( apply( dat[, grep(grp, names(dat))] , 1, function(z) sum(z, na.rm=TRUE)>=2) ) )
[[1]]
[1] 1 2 4 6 7
[[2]]
[1] 1 2 3 5 6 7
[[3]]
[1] 1 2 3 7
Now ,,,,,,, what exactly are the rutes for this task?????