Hmisc (version 4.3-0)

reShape: Reshape Matrices and Serial Data

Description

If the first argument is a matrix, reShape strings out its values and creates row and column vectors specifying the row and column each element came from. This is useful for sending matrices to Trellis functions, for analyzing or plotting results of table or crosstabs, or for reformatting serial data stored in a matrix (with rows representing multiple time points) into vectors. The number of observations in the new variables will be the product of the number of rows and number of columns in the input matrix. If the first argument is a vector, the id and colvar variables are used to restructure it into a matrix, with NAs for elements that corresponded to combinations of id and colvar values that did not exist in the data. When more than one vector is given, multiple matrices are created. This is useful for restructuring irregular serial data into regular matrices. It is also useful for converting data produced by expand.grid into a matrix (see the last example). The number of rows of the new matrices equals the number of unique values of id, and the number of columns equals the number of unique values of colvar.

When the first argument is a vector and the id is a data frame (even with only one variable), reShape will produce a data frame, and the unique groups are identified by combinations of the values of all variables in id. If a data frame constant is specified, the variables in this data frame are assumed to be constant within combinations of id variables (if not, an arbitrary observation in constant will be selected for each group). A row of constant corresponding to the target id combination is then carried along when creating the data frame result.

A different behavior of reShape is achieved when base and reps are specified. In that case x must be a list or data frame, and those data are assumed to contain one or more non-repeating measurements (e.g., baseline measurements) and one or more repeated measurements represented by variables named by pasting together the character strings in the vector base with the integers 1, 2, …, reps. The input data are rearranged by repeating each value of the baseline variables reps times and by transposing each observation's values of one of the set of repeated measurements as reps observations under the variable whose name does not have an integer pasted to the end. if x has a row.names attribute, those observation identifiers are each repeated reps times in the output object. See the last example.

Usage

reShape(x, …, id, colvar, base, reps, times=1:reps,
        timevar='seqno', constant=NULL)

Arguments

x

a matrix or vector, or, when base is specified, a list or data frame

other optional vectors, if x is a vector

id

A numeric, character, category, or factor variable containing subject identifiers, or a data frame of such variables that in combination form groups of interest. Required if x is a vector, ignored otherwise.

colvar

A numeric, character, category, or factor variable containing column identifiers. colvar is using a "time of data collection" variable. Required if x is a vector, ignored otherwise.

base

vector of character strings containing base names of repeated measurements

reps

number of times variables named in base are repeated. This must be a constant.

times

when base is given, times is the vector of times to create if you do not want to use consecutive integers beginning with 1.

timevar

specifies the name of the time variable to create if times is given, if you do not want to use seqno

constant

a data frame with the same number of rows in id and x, containing auxiliary information to be merged into the resulting data frame. Logically, the rows of constant within each group should have the same value of all of its variables.

Value

If x is a matrix, returns a list containing the row variable, the column variable, and the as.vector(x) vector, named the same as the calling argument was called for x. If x is a vector and no other vectors were specified as , the result is a matrix. If at least one vector was given to , the result is a list containing k matrices, where k one plus the number of vectors in . If x is a list or data frame, the same type of object is returned. If x is a vector and id is a data frame, a data frame will be the result.

Details

In converting dimnames to vectors, the resulting variables are numeric if all elements of the matrix dimnames can be converted to numeric, otherwise the corresponding row or column variable remains character. When the dimnames if x have a names attribute, those two names become the new variable names. If x is a vector and another vector is also given (in ), the matrices in the resulting list are named the same as the input vector calling arguments. You can specify customized names for these on-the-fly by using e.g. reShape(X=x, Y=y, id= , colvar= ). The new names will then be X and Y instead of x and y. A new variable named seqnno is also added to the resulting object. seqno indicates the sequential repeated measurement number. When base and times are specified, this new variable is named the character value of timevar and the values are given by a table lookup into the vector times.

See Also

reshape, as.vector, matrix, dimnames, outer, table

Examples

Run this code
# NOT RUN {
set.seed(1)
Solder  <- factor(sample(c('Thin','Thick'),200,TRUE),c('Thin','Thick'))
Opening <- factor(sample(c('S','M','L'),  200,TRUE),c('S','M','L'))

tab <- table(Opening, Solder)
tab
reShape(tab)
# attach(tab)  # do further processing

# An example where a matrix is created from irregular vectors
follow <- data.frame(id=c('a','a','b','b','b','d'),
                     month=c(1, 2,  1,  2,  3,  2),
                     cholesterol=c(225,226, 320,319,318, 270))
follow
attach(follow)
reShape(cholesterol, id=id, colvar=month)
detach('follow')
# Could have done :
# reShape(cholesterol, triglyceride=trig, id=id, colvar=month)

# Create a data frame, reshaping a long dataset in which groups are
# formed not just by subject id but by combinations of subject id and
# visit number.  Also carry forward a variable that is supposed to be
# constant within subject-visit number combinations.  In this example,
# it is not constant, so an arbitrary visit number will be selected.
w <- data.frame(id=c('a','a','a','a','b','b','b','d','d','d'),
             visit=c(  1,  1,  2,  2,  1,  1,  2,  2,  2,  2),
                 k=c('A','A','B','B','C','C','D','E','F','G'),
               var=c('x','y','x','y','x','y','y','x','y','z'),
               val=1:10)
with(w,
     reShape(val, id=data.frame(id,visit),
             constant=data.frame(k), colvar=var))

# Get predictions from a regression model for 2 systematically
# varying predictors.  Convert the predictions into a matrix, with
# rows corresponding to the predictor having the most values, and
# columns corresponding to the other predictor
# d <- expand.grid(x2=0:1, x1=1:100)
# pred <- predict(fit, d)
# reShape(pred, id=d$x1, colvar=d$x2)  # makes 100 x 2 matrix


# Reshape a wide data frame containing multiple variables representing
# repeated measurements (3 repeats on 2 variables; 4 subjects)
set.seed(33)
n <- 4
w <- data.frame(age=rnorm(n, 40, 10),
                sex=sample(c('female','male'), n,TRUE),
                sbp1=rnorm(n, 120, 15),
                sbp2=rnorm(n, 120, 15),
                sbp3=rnorm(n, 120, 15),
                dbp1=rnorm(n,  80, 15),
                dbp2=rnorm(n,  80, 15),
                dbp3=rnorm(n,  80, 15), row.names=letters[1:n])
options(digits=3)
w


u <- reShape(w, base=c('sbp','dbp'), reps=3)
u
reShape(w, base=c('sbp','dbp'), reps=3, timevar='week', times=c(0,3,12))
# }

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