R Basics

MATH-7360 Data Analysis

Computer storage

• bit = “binary” + “digit” (0 or 1)

• byte = 8 bits

• kB = kilobyte = \(10^3\) bytes

• MB = megabytes = \(10^6\) bytes

• GB = gigabytes = \(10^9\) bytes

• TB = terabytes = \(10^{12}\) bytes

• PB = petabytes = \(10^{15}\) bytes

Fixed-point number system

Fixed-point number system is a computer model for integers. One storage unit may be \(M = 8/16/32/64\) bit.

• The first bit stores the sign of the integer (i.e. 0 for positive and 1 for negative)

• The rest of the bits store the absolute value of the integer in binary

• Range of representable integers by \(M\)-bit storage unit is \([-2^{M - 1}, 2^{M - 1} - 1]\) (don’t need to represent \(0\) anymore so could have capacity for \(2^{M-1}\) negative numbers).

• For \(M=8\), \([-128, 127]\). For \(M=16\), \([-65536, 65535]\). For \(M=32\), \([-2147483648, 2147483647]\).

.Machine$integer.max

## [1] 2147483647

# integer type in R uses M=32 bits
M <- 32
big <- 2^(M-1) - 1
small <- -2^(M-1)
as.integer(big)

## [1] 2147483647

as.integer(big + 1)

## Warning: NAs introduced by coercion to integer range

## [1] NA

as.integer(small + 1)

## [1] -2147483647

as.integer(small)

## Warning: NAs introduced by coercion to integer range

## [1] NA

• R’s smallest integer is \(-2^{31} + 1 = -2147483647\).

Keep track of overflow and underflow. If the result of a summation is \(R\), which must be in the set \([-2^{M - 1}, 2^{M - 1} - 1]\), there are only three possibilities for the true sum: \(R\), \(R+2^M\) (overflow), or \(R-2^M\) (underflow).

Floating-point number system

Floating-point number system is a computer model for real numbers.

• A real is represented by \(\mbox{value} = (-1)^{b_{31}} \times 2^{(b_{30} b_{29} \dots b_{23})_2 - 127} \times (1.b_{22} b_{21} \dots b_{0})_2\).

• IEEE 754-1985 and IEEE 754-2008

  • Single precision (32 bit): base \(b = 2\), \(p=23\) (23 significant bits), \(e_{\mbox{max}}=127\), \(e_{\mbox{min}}=-126\) (8 exponent bits), \(\mbox{bias} = 127\). This implies a maximum magnitude of \(\log_{10}(2^{127}) \approx 38\) and precision to \(\log_{10}(2^{23}) \approx 7\) decimal point. \(\pm 10^{\pm38}\).

  • Double precision (64 bit): base \(b = 2\), \(p=52\) (52 significant bits), \(e_{\mbox{max}}=1023\), \(e_{\mbox{min}}=-1022\) (11 exponent bits), \(\mbox{bias} = 1023\). This implies a maximum magnitude of \(\log_{10}(2^{1023}) \approx 308\) and precision to \(\log_{10}(2^{52}) \approx 16\) decimal point. \(\pm 10^{\pm308}\).

  • In the above example, \(\mbox{value} =\)

2^((2^2 + 2^3 + 2^4 + 2^5 + 2^6) - 127) * (1 + 2^(-2))

## [1] 0.15625

• R only uses double (64-bit) and 32-bit integer. It can be a downside when dealing with big data.

To summarize

• Single precision: \(\pm 10^{\pm38}\) with precision up to \(7\) decimal digits.

• Double precision: \(\pm 10^{\pm308}\) with precision up to \(16\) decimal digits.

• The floating-point numbers do not occur uniformly over the real number line.

• The variable \(\mbox{.Machine}\) in R contains numerical characteristics of the machine.

• How to test \(\mbox{inf}\) and \(\mbox{nan}\)? In R