问题
Edit -- This question was originally titled << Long to wide data reshaping in R >>
I'm just learning R and trying to find ways to apply it to help out others in my life. As a test case, I'm working on reshaping some data, and I'm having trouble following the examples I've found online. What I'm starting with looks like this:
ID Obs 1 Obs 2 Obs 3
1 43 48 37
1 27 29 22
1 36 32 40
2 33 38 36
2 29 32 27
2 32 31 35
2 25 28 24
3 45 47 42
3 38 40 36
And what I want to end up with will look like this:
ID Obs 1 mean Obs 1 std dev Obs 2 mean Obs 2 std dev
1 x x x x
2 x x x x
3 x x x x
And so forth. What I'm unsure of is whether I need additional information in my long-form data, or what. I imagine that the math part (finding the mean and standard deviations) will be the easy part, but I haven't been able to find a way that seems to work to reshape the data correctly to start in on that process.
Thanks very much for any help.
回答1:
This is an aggregation problem, not a reshaping problem as the question originally suggested -- we wish to aggregate each column into a mean and standard deviation by ID. There are many packages that handle such problems. In the base of R it can be done using aggregate like this (assuming DF is the input data frame):
ag <- aggregate(. ~ ID, DF, function(x) c(mean = mean(x), sd = sd(x)))
Note 1: A commenter pointed out that ag is a data frame for which some columns are matrices. Although initially that may seem strange, in fact it simplifies access. ag has the same number of columns as the input DF. Its first column ag[[1]] is ID and the ith column of the remainder ag[[i+1]] (or equivalanetly ag[-1][[i]]) is the matrix of statistics for the ith input observation column. If one wishes to access the jth statistic of the ith observation it is therefore ag[[i+1]][, j] which can also be written as ag[-1][[i]][, j] .
On the other hand, suppose there are k statistic columns for each observation in the input (where k=2 in the question). Then if we flatten the output then to access the jth statistic of the ith observation column we must use the more complex ag[[k*(i-1)+j+1]] or equivalently ag[-1][[k*(i-1)+j]] .
For example, compare the simplicity of the first expression vs. the second:
ag[-1][[2]]
## mean sd
## [1,] 36.333 10.2144
## [2,] 32.250 4.1932
## [3,] 43.500 4.9497
ag_flat <- do.call("data.frame", ag) # flatten
ag_flat[-1][, 2 * (2-1) + 1:2]
## Obs_2.mean Obs_2.sd
## 1 36.333 10.2144
## 2 32.250 4.1932
## 3 43.500 4.9497
Note 2: The input in reproducible form is:
Lines <- "ID Obs_1 Obs_2 Obs_3
1 43 48 37
1 27 29 22
1 36 32 40
2 33 38 36
2 29 32 27
2 32 31 35
2 25 28 24
3 45 47 42
3 38 40 36"
DF <- read.table(text = Lines, header = TRUE)
回答2:
There are a few different ways to go about it. reshape2 is a helpful package.
Personally, I like using data.table
Below is a step-by-step
If myDF is your data.frame:
library(data.table)
DT <- data.table(myDF)
DT
# this will get you your mean and SD's for each column
DT[, sapply(.SD, function(x) list(mean=mean(x), sd=sd(x)))]
# adding a `by` argument will give you the groupings
DT[, sapply(.SD, function(x) list(mean=mean(x), sd=sd(x))), by=ID]
# If you would like to round the values:
DT[, sapply(.SD, function(x) list(mean=round(mean(x), 3), sd=round(sd(x), 3))), by=ID]
# If we want to add names to the columns
wide <- setnames(DT[, sapply(.SD, function(x) list(mean=round(mean(x), 3), sd=round(sd(x), 3))), by=ID], c("ID", sapply(names(DT)[-1], paste0, c(".men", ".SD"))))
wide
ID Obs.1.men Obs.1.SD Obs.2.men Obs.2.SD Obs.3.men Obs.3.SD
1: 1 35.333 8.021 36.333 10.214 33.0 9.644
2: 2 29.750 3.594 32.250 4.193 30.5 5.916
3: 3 41.500 4.950 43.500 4.950 39.0 4.243
Also, this may or may not be helpful
> DT[, sapply(.SD, summary), .SDcols=names(DT)[-1]]
Obs.1 Obs.2 Obs.3
Min. 25.00 28.00 22.00
1st Qu. 29.00 31.00 27.00
Median 33.00 32.00 36.00
Mean 34.22 36.11 33.22
3rd Qu. 38.00 40.00 37.00
Max. 45.00 48.00 42.00
回答3:
Here is probably the simplest way to go about it (with a reproducible example):
library(plyr)
df <- data.frame(ID=rep(1:3, 3), Obs_1=rnorm(9), Obs_2=rnorm(9), Obs_3=rnorm(9))
ddply(df, .(ID), summarize, Obs_1_mean=mean(Obs_1), Obs_1_std_dev=sd(Obs_1),
Obs_2_mean=mean(Obs_2), Obs_2_std_dev=sd(Obs_2))
ID Obs_1_mean Obs_1_std_dev Obs_2_mean Obs_2_std_dev
1 1 -0.13994642 0.8258445 -0.15186380 0.4251405
2 2 1.49982393 0.2282299 0.50816036 0.5812907
3 3 -0.09269806 0.6115075 -0.01943867 1.3348792
EDIT: The following approach saves you a lot of typing when dealing with many columns.
ddply(df, .(ID), colwise(mean))
ID Obs_1 Obs_2 Obs_3
1 1 -0.3748831 0.1787371 1.0749142
2 2 -1.0363973 0.0157575 -0.8826969
3 3 1.0721708 -1.1339571 -0.5983944
ddply(df, .(ID), colwise(sd))
ID Obs_1 Obs_2 Obs_3
1 1 0.8732498 0.4853133 0.5945867
2 2 0.2978193 1.0451626 0.5235572
3 3 0.4796820 0.7563216 1.4404602
回答4:
I add the dplyr solution.
set.seed(1)
df <- data.frame(ID=rep(1:3, 3), Obs_1=rnorm(9), Obs_2=rnorm(9), Obs_3=rnorm(9))
library(dplyr)
df %>% group_by(ID) %>% summarise_each(funs(mean, sd))
# ID Obs_1_mean Obs_2_mean Obs_3_mean Obs_1_sd Obs_2_sd Obs_3_sd
# (int) (dbl) (dbl) (dbl) (dbl) (dbl) (dbl)
# 1 1 0.4854187 -0.3238542 0.7410611 1.1108687 0.2885969 0.1067961
# 2 2 0.4171586 -0.2397030 0.2041125 0.2875411 1.8732682 0.3438338
# 3 3 -0.3601052 0.8195368 -0.4087233 0.8105370 0.3829833 1.4705692
回答5:
Here's another take on the data.table answers, using @Carson's data, that's a bit more readable (and also a little faster, because of using lapply instead of sapply):
library(data.table)
set.seed(1)
dt = data.table(ID=c(1:3), Obs_1=rnorm(9), Obs_2=rnorm(9), Obs_3=rnorm(9))
dt[, c(mean = lapply(.SD, mean), sd = lapply(.SD, sd)), by = ID]
# ID mean.Obs_1 mean.Obs_2 mean.Obs_3 sd.Obs_1 sd.Obs_2 sd.Obs_3
#1: 1 0.4854187 -0.3238542 0.7410611 1.1108687 0.2885969 0.1067961
#2: 2 0.4171586 -0.2397030 0.2041125 0.2875411 1.8732682 0.3438338
#3: 3 -0.3601052 0.8195368 -0.4087233 0.8105370 0.3829833 1.4705692
来源：https://stackoverflow.com/questions/16367436/compute-mean-and-standard-deviation-by-group-for-multiple-variables-in-a-data-fr