Skip to contents

Calculate eGFR Using CKD-EPI 2009 Equation

ckdepi_2009_egfr.Rd

Calculate eGFR Using CKD-EPI 2009 Equation

Usage

ckdepi_2009_egfr(sexf, raceb, age, creat)

Arguments

sexf

boolean value of sex Female: TRUE, Male: FALSE

raceb

boolean value of Race == Black: Black: TRUE, Other: FALSE

age

age of subject (years)

creat

creatinine levels of subject (mg/dL)

Value

the eGFR value (mL/min/1.73m2)

Details

The CKD-EPI 2009 equation: $$eGFR = 141 \cdot \min(S_{cr}/\kappa, 1)^\alpha \cdot \max(S_{cr}/\kappa, 1)^{-1.209} \cdot 0.993^A \cdot 1.018^F \cdot 1.159^B$$

where:

  • \(S_{cr}\) = serum creatinine (mg/dL)

  • \(\kappa\) = 0.7 (female) or 0.9 (male)

  • \(\alpha\) = -0.329 (female) or -0.411 (male)

  • \(A\) = age (years)

  • \(F\) = 1 (female) or 0 (male)

  • \(B\) = 1 (Black) or 0 (other)

See also

Other renal_function: aegfr(), ckdepi_2021_egfr(), ckdepi_2021_egfr_cystatin(), crcl(), egfr(), mdrd_egfr(), rfc(), schwartz_egfr()

Examples

e <- ckdepi_2009_egfr(TRUE, TRUE, 24, 1)

df <- data.frame(
  "SEXF" = c(TRUE, FALSE, TRUE, FALSE),
  "RACEB" = c(FALSE, FALSE, TRUE, FALSE),
  "AGE" = c(24, 24, 23, 24),
  "CREAT" = c(1, 1, 2, 1)
)
df <- dplyr::mutate(df, egfr = ckdepi_2009_egfr(SEXF, RACEB, AGE, CREAT))
df
#>    SEXF RACEB AGE CREAT      egfr
#> 1  TRUE FALSE  24     1  78.79023
#> 2 FALSE FALSE  24     1 104.87700
#> 3  TRUE  TRUE  23     2  39.77968
#> 4 FALSE FALSE  24     1 104.87700