Reduce0exact {SSBtools}

Reducing a non-negative regression problem

Description

The linear equation problem, z = t(x) %*% y with y non-negative and x as a design (dummy) matrix, is reduced to a smaller problem by identifying elements of y that can be found exactly from x and z.

Usage

Reduce0exact(
  x,
  z = NULL,
  reduceByColSums = FALSE,
  reduceByLeverage = FALSE,
  leverageLimit = 0.999999,
  digitsRoundWhole = 9,
  y = NULL,
  yStart = NULL,
  printInc = FALSE
)

Arguments

x

A matrix

z

A single column matrix

reduceByColSums

See Details

reduceByLeverage

See Details

leverageLimit

Limit to determine perfect fit

digitsRoundWhole

RoundWhole parameter for fitted values (when leverageLimit and y not in input)

y

A single column matrix. With y in input, z in input can be omitted and estimating y (when leverageLimit) is avoided.

yStart

A starting estimate when this function is combined with iterative proportional fitting. Zeros in yStart will be used to reduce the problem.

printInc

Printing iteration information to console when TRUE

Details

Exact elements can be identified in three ways in an iterative manner:

1. By zeros in z. This is always done.

2. By columns in x with a singe nonzero value. Done when reduceByColSums or reduceByLeverage is TRUE.

3. By exact linear regression fit (when leverage is one). Done when reduceByLeverage is TRUE. The leverages are computed by hat(as.matrix(x), intercept = FALSE), which can be very time and memory consuming. Furthermore, without y in input, known values will be computed by ginv.

Value

A list of five elements:

• x: A reduced version of input x

• z: Corresponding reduced z

• yKnown: Logical, specifying known values of y

• y: A version of y with known values correct and others zero

• zSkipped: Logical, specifying omitted columns of x

Author(s)

Øyvind Langsrud

Examples

# Make a special data set
d <- SSBtoolsData("sprt_emp")
d$ths_per <- round(d$ths_per)
d <- rbind(d, d)
d$year <- as.character(rep(2014:2019, each = 6))
to0 <- rep(TRUE, 36)
to0[c(6, 14, 17, 18, 25, 27, 30, 34, 36)] <- FALSE
d$ths_per[to0] <- 0

# Values as a single column matrix
y <- Matrix(d$ths_per, ncol = 1)

# A model matrix using a special year hierarchy
x <- Hierarchies2ModelMatrix(d, hierarchies = list(geo = "", age = "", year = 
    c("y1418 = 2014+2015+2016+2017+2018", "y1519 = 2015+2016+2017+2018+2019", 
      "y151719 = 2015+2017+2019", "yTotal = 2014+2015+2016+2017+2018+2019")), 
      inputInOutput = FALSE)

# Aggregates 
z <- t(x) %*% y
sum(z == 0)  # 5 zeros

# From zeros in z
a <- Reduce0exact(x, z)
sum(a$yKnown)   # 17 zeros in y is known
dim(a$x)        # Reduced x, without known y and z with zeros 
dim(a$z)        # Corresponding reduced z 
sum(a$zSkipped) # 5 elements skipped 
t(a$y)          # Just zeros (known are 0 and unknown set to 0) 

# It seems that three additional y-values can be found directly from z
sum(colSums(a$x) == 1)

# But it is the same element of y (row 18)
a$x[18, colSums(a$x) == 1]

# Make use of ones in colSums
a2 <- Reduce0exact(x, z, reduceByColSums = TRUE)
sum(a2$yKnown)          # 18 values in y is known
dim(a2$x)               # Reduced x
dim(a2$z)               # Corresponding reduced z
a2$y[which(a2$yKnown)]  # The known values of y (unknown set to 0)

# Six ones in leverage values 
# Thus six extra elements in y can be found by linear estimation
hat(as.matrix(a2$x), intercept = FALSE)

# Make use of ones in leverages (hat-values)
a3 <- Reduce0exact(x, z, reduceByLeverage = TRUE)
sum(a3$yKnown)          # 26 values in y is known (more than 6 extra)
dim(a3$x)               # Reduced x
dim(a3$z)               # Corresponding reduced z
a3$y[which(a3$yKnown)]  # The known values of y (unknown set to 0)

# More than 6 extra is caused by iteration 
# Extra checking of zeros in z after reduction by leverages 
# Similar checking performed also after reduction by colSums