I would like to ask how can I rearrange my dataset that fulfils the following
[
Original :
Group Value_y Value_z
1 m a
1 n a
2 o b
2 p b
Intended:
Group Value_a Value_b
1 m n
2 o p
]1
which involves separating value_y according to value_z and adding a new column according to the group number. Will potential need to average a separate column's values and add as a new column the same way.
Thank you!
In data.table we can use dcast :
library(data.table)
dcast(setDT(df), Group~rowid(Value_z), value.var = 'Value_y')
# Group 1 2
#1: 1 m n
#2: 2 o p
data
df <- structure(list(Group = c(1L, 1L, 2L, 2L), Value_y = c("m", "n",
"o", "p"), Value_z = c("a", "a", "b", "b")), class = "data.frame",
row.names = c(NA, -4L))
There is a dplyr solution. Define
Uneven = seq(1, dim(A)[1] - 1, by = 2)
Even = seq(2, dim(A)[1], by = 2)
with
A = data.frame(Group = c(1, 1, 2, 2), Value_y = c("m", "n", "o", "p"))
Then, you can use the pipe and some dplyr functionality to get
A2 = A %>%
dplyr::group_by(Group) %>%
dplyr::mutate(Row_1 = Value_y[Uneven]) %>%
dplyr::mutate(Row_2 = Value_y[Even]) %>%
dplyr::select(-Value_y) %>%
dplryr::slice(1)
and the output
> A2
# A tibble: 2 x 3
# Groups: Group [2]
Group Row_1 Row_2
<dbl> <fct> <fct>
1 1 m n
2 2 o p
Note that this solution presupposes two-pairs of Groups, i.e. an even number of observations.
Related
I want to fill df2 with information from df1.
df1 as below
ID Mutation
1 A
2 B
2 C
3 A
df2 as below
ID A B C
1
2
3
For example, if mutation A is found in ID 1, then I want it in df2 it marked as "Y".
So the df2 result should be
ID A B C
1 Y
2 Y Y
3 Y
I have hundreds of IDs and more than 20 mutations. How can I efficiently achieve this in R? Thanks!
Using data.table you can try
setDT(df)
df2 <- dcast(df,formula = ID~Mutation )
df2[, c("A", "B", "C") := lapply(.SD, function(x) ifelse(is.na(x), " ", "Y")), ID]
df2
#Output
ID A B C
1: 1 Y
2: 2 Y Y
3: 3 Y
Create a new column with value 'Y' and cast the data in wide format.
library(dplyr)
library(tidyr)
df %>%
mutate(value = 'Y') %>%
pivot_wider(names_from = Mutation, values_from = value, values_fill = '')
# ID A B C
# <int> <chr> <chr> <chr>
#1 1 "Y" "" ""
#2 2 "" "Y" "Y"
#3 3 "Y" "" ""
data
df <- structure(list(ID = c(1L, 2L, 2L, 3L), Mutation = c("A", "B",
"C", "A")), class = "data.frame", row.names = c(NA, -4L))
I have a df like
ProjectID Dist
1 x
1 y
2 z
2 x
2 h
3 k
.... ....
and a vector of indices of lengthunique(df$ProjectID) like
2
3
1
....
I would like to get Dist by ProjectID whose index is the element vector corresponding to project ID. So the result I want looks like
ProjectID Dist
1 y
2 h
3 k
.... ....
I tried
aggregate(XRKL ~ ID, FUN=..?, data=df)
but I'm not sure where I can put the vector of indices. Is there a way to get the right result from dply ftns, tapply, or aggregate? Or do I need to make a function of my own? Thank you.
You can add the indices in the dataframe itself and then select that row from each group.
inds <- c(2, 3, 1)
df %>%
mutate(inds = inds[match(ProjectID, unique(ProjectID))]) %>%
#If ProjectID is sequential like 1, 2, 3
#mutate(inds = inds[ProjectID]) %>%
group_by(ProjectID) %>%
slice(first(inds)) %>%
ungroup() %>%
select(-inds)
# ProjectID Dist
# <int> <chr>
#1 1 y
#2 2 h
#3 3 k
data
df <- structure(list(ProjectID = c(1L, 1L, 2L, 2L, 2L, 3L), Dist = c("x",
"y", "z", "x", "h", "k")), class = "data.frame", row.names = c(NA, -6L))
I imagine this is already solved in many places, but I lack the right wordage to use to search for a solution. In R I have example data in long format like this:
A = tibble( c(1,2,3,1,2,4,5,5), c('a','b','c','a','f','-','b', 'f'))
and what I want returned is sort of a grouped result (something like a spread?) where I first collect the set of letters that match each number to get something like this.
1: 'a', 'a'
2: 'b', 'f'
3: 'c', 'c'
4: '_'
5: 'b', 'f'
and the actual final result I am looking for is the count of how many times each letter combination, when is observed:
'a','a': 1
'b','f': 2
'c','c': 1
'-': 1
I can do the last step with group_by() but I mention it here in case there is some magic sauce that does the whole thing.
We can do a group by 'a', then paste the second column while taking the number of distinct elements in 'b' and get the distinct rows
library(dplyr)
library(stringr)
A %>%
group_by(a) %>%
summarise(out = str_c(b, collapse=","), n = n_distinct(b))%>%
distinct(out, n)
# A tibble: 4 x 2
# out n
# <chr> <int>
#1 a,a 1
#2 b,f 2
#3 c 1
#4 - 1
data
A <- structure(list(a = c(1, 2, 3, 1, 2, 4, 5, 5), b = c("a", "b",
"c", "a", "f", "-", "b", "f")), row.names = c(NA, -8L), class = c("tbl_df",
"tbl", "data.frame"))
This is close to what you are looking for:
library(tidyverse)
#Data
A <- structure(list(v1 = c(1, 2, 3, 1, 2, 4, 5, 5), v2 = c("a", "b",
"c", "a", "f", "-", "b", "f")), row.names = c(NA, -8L), class = c("tbl_df",
"tbl", "data.frame"))
#Code
A %>% group_by(v1) %>% summarise(chain=paste0(v2,collapse = ',')) %>% ungroup() %>%
group_by(chain) %>% summarise(N=n())
# A tibble: 4 x 2
chain N
<chr> <int>
1 - 1
2 a,a 1
3 b,f 2
4 c 1
Here is a base R option using nested aggregate
aggregate(.~y,aggregate(y~.,A,toString),length)
which gives
> aggregate(.~y,aggregate(y~.,A,toString),length)
y x
1 - 1
2 a, a 1
3 b, f 2
4 c 1
Data
A = tibble(x = c(1,2,3,1,2,4,5,5), y = c('a','b','c','a','f','-','b', 'f'))
Maybe you want to cast the data in wide format and then count the combinations. Try :
library(dplyr)
library(tidyr)
A %>%
group_by(v1) %>%
mutate(row = row_number()) %>%
pivot_wider(names_from = row, values_from = v2, names_prefix = 'col_') %>%
ungroup %>%
count(col_1, col_2)
# col_1 col_2 n
# <chr> <chr> <int>
#1 - NA 1
#2 a a 1
#3 b f 2
#4 c NA 1
I have two data frames. dfOne is made like this:
X Y Z T J
3 4 5 6 1
1 2 3 4 1
5 1 2 5 1
and dfTwo is made like this
C.1 C.2
X Z
Y T
I want to obtain a new dataframe where there are simultaneously X, Y, Z, T Values which are major than a specific threshold.
Example. I need simultaneously (in the same row):
X, Y > 2
Z, T > 4
I need to use the second data frame to reach my objective, I expect something like:
dfTwo$C.1>2
so the result would be a new dataframe with this structure:
X Y Z T J
3 4 5 6 1
How could I do it?
Here is a base R method with Map and Reduce.
# build lookup table of thresholds relative to variable name
vals <- setNames(c(2, 2, 4, 4), unlist(dat2))
# subset data.frame
dat[Reduce("&", Map(">", dat[names(vals)], vals)), ]
X Y Z T J
1 3 4 5 6 1
Here, Map returns a list of length 4 with logical variables corresponding to each comparison. This list is passed to Reduce which returns a single logical vector with length corresponding to the number of rows in the data.frame, dat. This logical vector is used to subset dat.
data
dat <-
structure(list(X = c(3L, 1L, 5L), Y = c(4L, 2L, 1L), Z = c(5L,
3L, 2L), T = c(6L, 4L, 5L), J = c(1L, 1L, 1L)), .Names = c("X",
"Y", "Z", "T", "J"), class = "data.frame", row.names = c(NA,
-3L))
dat2 <-
structure(list(C.1 = structure(1:2, .Label = c("X", "Y"), class = "factor"),
C.2 = structure(c(2L, 1L), .Label = c("T", "Z"), class = "factor")), .Names = c("C.1",
"C.2"), class = "data.frame", row.names = c(NA, -2L))
We can use the purrr package
Here is the input data.
# Data frame from lmo's solution
dat <-
structure(list(X = c(3L, 1L, 5L), Y = c(4L, 2L, 1L), Z = c(5L,
3L, 2L), T = c(6L, 4L, 5L), J = c(1L, 1L, 1L)), .Names = c("X",
"Y", "Z", "T", "J"), class = "data.frame", row.names = c(NA,
-3L))
# A numeric vector to show the threshold values
# Notice that columns without any requirements need NA
vals <- c(X = 2, Y = 2, Z = 4, T = 4, J = NA)
Here is the implementation
library(purrr)
map2_dfc(dat, vals, ~ifelse(.x > .y | is.na(.y), .x, NA)) %>% na.omit()
# A tibble: 1 x 5
X Y Z T J
<int> <int> <int> <int> <int>
1 3 4 5 6 1
map2_dfc loop through each column in dat and each value in vals one by one with a defined function. ~ifelse(.x > .y | is.na(.y), .x, NA) means if the number in each column is larger than the corresponding value in vals, or vals is NA, the output should be the original value from the column. Otherwise, the value is replaced to be NA. The output of map2_dfc(dat, vals, ~ifelse(.x > .y | is.na(.y), .x, NA)) is a data frame with NA values in some rows indicating that the condition is not met. Finally, na.omit removes those rows.
Update
Here I demonstrate how to covert the dfTwo dataframe to the vals vector in my example.
First, let's create the dfTwo data frame.
dfTwo <- read.table(text = "C.1 C.2
X Z
Y T",
header = TRUE, stringsAsFactors = FALSE)
dfTwo
C.1 C.2
1 X Z
2 Y T
To complete the task, I load the dplyr and tidyr package.
library(dplyr)
library(tidyr)
Now I begin the transformation of dfTwo. The first step is to use stack function to convert the format.
dfTwo2 <- dfTwo %>%
stack() %>%
setNames(c("Col", "Group")) %>%
mutate(Group = as.character(Group))
dfTwo2
Col Group
1 X C.1
2 Y C.1
3 Z C.2
4 T C.2
The second step is to add the threshold information. One way to do this is to create a look-up table showing the association between Group and Value
threshold_df <- data.frame(Group = c("C.1", "C.2"),
Value = c(2, 4),
stringsAsFactors = FALSE)
threshold_df
Group Value
1 C.1 2
2 C.2 4
And then we can use the left_join function to combine the data frame.
dfTwo3 <- dfTwo2 %>% left_join(threshold_dt, by = "Group")
dfTwo3
Col Group Value
1 X C.1 2
2 Y C.1 2
3 Z C.2 4
4 T C.2 4
Now it is the third step. Notice that there is a column called J which does not need any threshold. So we need to add this information to dfTwo3. We can use the complete function from tidyr. The following code completes the data frame by adding Col in dat but not in dfTwo3 and NA to the Value.
dfTwo4 <- dfTwo3 %>% complete(Col = colnames(dat))
dfTwo4
# A tibble: 5 x 3
Col Group Value
<chr> <chr> <dbl>
1 J <NA> NA
2 T C.2 4
3 X C.1 2
4 Y C.1 2
5 Z C.2 4
The fourth step is arrange the right order of dfTwo4. We can achieve this by turning Col to factor and assign the level based on the order of the column name in dat.
dfTwo5 <- dfTwo4 %>%
mutate(Col = factor(Col, levels = colnames(dat))) %>%
arrange(Col) %>%
mutate(Col = as.character(Col))
dfTwo5
# A tibble: 5 x 3
Col Group Value
<chr> <chr> <dbl>
1 X C.1 2
2 Y C.1 2
3 Z C.2 4
4 T C.2 4
5 J <NA> NA
We are almost there. Now we can create vals from dfTwo5.
vals <- dfTwo5$Value
names(vals) <- dfTwo5$Col
vals
X Y Z T J
2 2 4 4 NA
Now we are ready to use the purrr package to filter the data.
The aboved are the breakdown of steps. We can combine all these steps into the following code for simlicity.
library(dplyr)
library(tidyr)
threshold_df <- data.frame(Group = c("C.1", "C.2"),
Value = c(2, 4),
stringsAsFactors = FALSE)
dfTwo2 <- dfTwo %>%
stack() %>%
setNames(c("Col", "Group")) %>%
mutate(Group = as.character(Group)) %>%
left_join(threshold_df, by = "Group") %>%
complete(Col = colnames(dat)) %>%
mutate(Col = factor(Col, levels = colnames(dat))) %>%
arrange(Col) %>%
mutate(Col = as.character(Col))
vals <- dfTwo2$Value
names(vals) <- dfTwo2$Col
dfOne[Reduce(intersect, list(which(dfOne["X"] > 2),
which(dfOne["Y"] > 2),
which(dfOne["Z"] > 4),
which(dfOne["T"] > 4))),]
# X Y Z T J
#1 3 4 5 6 1
Or iteratively (so fewer inequalities are tested):
vals = c(X = 2, Y = 2, Z = 4, T = 4) # from #lmo's answer
dfOne[Reduce(intersect, lapply(names(vals), function(x) which(dfOne[x] > vals[x]))),]
# X Y Z T J
#1 3 4 5 6 1
I'm writing this assuming that the second DF is meant to categorize the fields in the first DF. It's way simpler if you don't need to use the second one to define the conditions:
dfNew = dfOne[dfOne$X > 2 & dfOne$Y > 2 & dfOne$Z > 4 & dfOne$T > 4, ]
Or, using dplyr:
library(dplyr)
dfNew = dfOne %>% filter(X > 2 & Y > 2 & Z > 4 & T > 4)
In case that's all you need, I'll save this comment while I poke at the more complicated version of the question.
This question already has answers here:
Create unique identifier from the interchangeable combination of two variables
(2 answers)
Closed 6 years ago.
I have a dataframe of 3 columns
A B 1
A B 1
A C 1
B A 1
I want to aggregate it such that it considers combinations A-B and B-A to be the same, resulting in
A B 3
A C 1
How do I go about this?
Use pmin and pmax on the first two columns and then do the group-by-count:
library(dplyr);
df %>% group_by(G1 = pmin(V1, V2), G2 = pmax(V1, V2)) %>% summarise(Count = sum(V3))
Source: local data frame [2 x 3]
Groups: G1 [?]
G1 G2 Count
(chr) (chr) (int)
1 A B 3
2 A C 1
Corresponding data.table solution would be:
library(data.table)
setDT(df)
df[, .(Count = sum(V3)), .(G1 = pmin(V1, V2), G2 = pmax(V1, V2))]
G1 G2 Count
1: A B 3
2: A C 1
Data:
structure(list(V1 = c("A", "A", "A", "B"), V2 = c("B", "B", "C",
"A"), V3 = c(1L, 1L, 1L, 1L)), .Names = c("V1", "V2", "V3"), row.names = c(NA,
-4L), class = "data.frame")