Introduction
Part of the work of Martin et al.
(2022) required transforming blood pressure measurement into
percentiles based on published norms. This work was complicated by the
fact that data for pediatric blood pressure percentiles is sparse and
generally only applicable to children at least one year of age and
requires height, a commonly unavailable data point in electronic health
records for a variety of reasons.
A solution to building pediatric blood pressure percentiles was
developed and is presented here for others to use. Inputs for the
developed method are:
- Patient sex (male/female) required
- Systolic blood pressure (mmHg) required
- Diastolic blood pressure (mmHg) required
- Patient height (cm) if known.
Given the inputs, the following logic is used to determine which data
sets will be used to inform the blood pressure percentiles. Under one
year of age, the data from Gemelli et al.
(1990) will be used; a height input is not required for this
patient subset. For those at least one year of age with a known height,
data from (nhlbi2011exper?) (hereafter
referred to as ‘NHLBI/CDC’ as the report incorporates recommendations
and inputs from the National Heart, Lung, and Blood Institute [NHLBI]
and the Centers for Disease Control and Prevention [CDC]). If height is
unknown and age is at least three years, then data from Lo et al. (2013) is used. Lastly, for children
between one and three years of age with unknown height, blood pressure
percentiles are estimated by the NHLBI/CDC data using as a default the
median height for each patient’s sex and age.
Estimating Pediatric
Blood Pressure Distributions
There are two functions provided for working with blood pressure
distributions. These methods use Gaussian distributions for both
systolic and diastolic blood pressures with means and standard
deviations either explicitly provided in an aforementioned source or
derived by optimizing the parameters such that the sum of squared errors
between the provided quantiles from an aforementioned source and the
distribution quantiles is minimized. The provided functions, a
distribution function and a quantile function, follow a similar naming
convention to the distribution functions found in the stats library in
R.
args(p_bp)
## function (q_sbp, q_dbp, age, male, height = NA, height_percentile = 0.5,
## ...)
## NULL
# Quantile Function
args(q_bp)
## function (p_sbp, p_dbp, age, male, height = NA, height_percentile = 0.5,
## ...)
## NULL
Both methods expect an age in months and an indicator for sex. If
height is missing, e.g., NA, then the default height percentile of 50
will be used as applicable based on the patient’s age group. The end
user may modify the default height percentile.
If height is entered, then the height percentile is determined via an
LMS method for age and sex using corresponding LMS data from the CDC
(more information on LMS methods and data is provided later in this
vignette). The parameters for the blood pressure distribution are found
in a look up table using the nearest age and height percentile.
Percentiles
What percentile for systolic and diastolic blood pressure is 100/60
for a 44 month old male with unknown height?
p_bp(q_sbp = 100, q_dbp = 60, age = 44, male = 1)
## $sbp_percentile
## [1] 0.7700861
##
## $dbp_percentile
## [1] 0.72739
Those percentiles would be modified if height was 183 cm:
p_bp(q_sbp = 100, q_dbp = 60, age = 44, male = 1, height = 183)
## $sbp_percentile
## [1] 0.676432
##
## $dbp_percentile
## [1] 0.8485735
The package can also be used to determine the blood pressure
percentiles corresponding to a child of a given height percentile. First
find the height quantile using the q_stature_for_age function, and then
use this height measurement (provided in centimeters) as the height
input for the p_bp function.
ht <- q_stature_for_age(p = 0.90, age = 44, male = 1)
ht
## [1] 105.2123
p_bp(q_sbp = 100, q_dbp = 60, age = 44, male = 1, height = ht)
## $sbp_percentile
## [1] 0.7086071
##
## $dbp_percentile
## [1] 0.8485735
A plotting method to show where the observed blood pressures are on
the distribution function is also provided.
bp_cdf(age = 44, male = 1, height = ht, sbp = 100, dbp = 60)
Vectors of blood pressures can be used as well. NA values will return
NA.
bps <-
p_bp(
q_sbp = c(100, NA, 90)
, q_dbp = c(60, 82, 48)
, age = 44
, male = 1
, height = ht
)
bps
## $sbp_percentile
## [1] 0.7086071 NA 0.3570545
##
## $dbp_percentile
## [1] 0.8485735 0.9982486 0.4998928
If you want to know which data source was used in computing each of
the percentile estimates you can look at the bp_params attribute:
attr(bps, "bp_params")
## source male age sbp_mean sbp_sd dbp_mean dbp_sd height_percentile
## 147 nhlbi 1 36 94.00085 10.92104 48.00313 11.64367 90
## 1471 nhlbi 1 36 94.00085 10.92104 48.00313 11.64367 90
## 1472 nhlbi 1 36 94.00085 10.92104 48.00313 11.64367 90
str(bps)
## List of 2
## $ sbp_percentile: num [1:3] 0.709 NA 0.357
## $ dbp_percentile: num [1:3] 0.849 0.998 0.5
## - attr(*, "bp_params")='data.frame': 3 obs. of 8 variables:
## ..$ source : chr [1:3] "nhlbi" "nhlbi" "nhlbi"
## ..$ male : int [1:3] 1 1 1
## ..$ age : num [1:3] 36 36 36
## ..$ sbp_mean : num [1:3] 94 94 94
## ..$ sbp_sd : num [1:3] 10.9 10.9 10.9
## ..$ dbp_mean : num [1:3] 48 48 48
## ..$ dbp_sd : num [1:3] 11.6 11.6 11.6
## ..$ height_percentile: int [1:3] 90 90 90
## - attr(*, "class")= chr "pedbp_bp"
Quantiles
If you have a percentile value and want to know the associated
systolic and diastolic blood pressures:
q_bp(
p_sbp = c(0.701, NA, 0.36)
, p_dbp = c(0.85, 0.99, 0.50)
, age = 44
, male = 1
, height = ht
)
## $sbp
## [1] 99.75929 NA 90.08611
##
## $dbp
## [1] 60.07101 75.09035 48.00313
Working With More
Than One Patient
The p_bp and q_bp methods are designed accept vectors for each of the
arguments. These methods expected each argument to be length 1 or all
the same length.
eg_data <- read.csv(system.file("example_data", "for_batch.csv", package = "pedbp"))
eg_data
## pid age_months male height..cm. sbp..mmHg. dbp..mmHg.
## 1 patient_A 96 1 NA 102 58
## 2 patient_B 144 0 153 113 NA
## 3 patient_C 4 0 62 82 43
## 4 patient_D_1 41 1 NA 96 62
## 5 patient_D_2 41 1 101 96 62
bp_percentiles <-
p_bp(
q_sbp = eg_data$sbp..mmHg.
, q_dbp = eg_data$dbp..mmHg.
, age = eg_data$age
, male = eg_data$male
, height = eg_data$height
)
bp_percentiles
## $sbp_percentile
## [1] 0.5533069 0.7680548 0.2622697 0.6195685 0.6101926
##
## $dbp_percentile
## [1] 0.4120704 NA 0.1356661 0.8028518 0.9011263
str(bp_percentiles)
## List of 2
## $ sbp_percentile: num [1:5] 0.553 0.768 0.262 0.62 0.61
## $ dbp_percentile: num [1:5] 0.412 NA 0.136 0.803 0.901
## - attr(*, "bp_params")='data.frame': 5 obs. of 8 variables:
## ..$ source : chr [1:5] "lo2013" "nhlbi" "gemelli1990" "lo2013" ...
## ..$ male : int [1:5] 1 0 0 1 1
## ..$ age : num [1:5] 96 144 3 36 36
## ..$ sbp_mean : num [1:5] 100.7 105 89 93.2 93
## ..$ sbp_sd : num [1:5] 9.7 10.9 11 9.2 10.7
## ..$ dbp_mean : num [1:5] 59.8 62 54 55.1 47
## ..$ dbp_sd : num [1:5] 8.1 10.9 10 8.1 11.6
## ..$ height_percentile: int [1:5] NA 50 NA NA 75
## - attr(*, "class")= chr "pedbp_bp"
Going from percentiles back to quantiles:
q_bp(
p_sbp = bp_percentiles$sbp_percentile
, p_dbp = bp_percentiles$dbp_percentile
, age = eg_data$age
, male = eg_data$male
, height = eg_data$height
)
## $sbp
## [1] 102 113 82 96 96
##
## $dbp
## [1] 58 NA 43 62 62
Shiny Application
An interactive Shiny
application is also available. After installing the pedbp package and
the suggested packages, you can run the app locally via
shiny::runApp(system.file("shinyapps", "pedbp", package = "pedbp"))
The shiny application allows for interactive exploration of blood
pressure percentiles for an individual patient and allows for batch
processing a set of patients as well.
An example input file for batch processing is provided within the
package an can be accessed via:
system.file("example_data", "for_batch.csv", package = "pedbp")
CDC Growth Charts
Using the Percentile
Data Files with LMS values provided by the CDC, we provide eight
distribution tools:
- weight for age for infants
- length for age for infants
- weight for length for infants
- head circumference for age
- weight for stature
- weight for age
- stature for age
- BMI for age
All lengths/heights are in centimeters, ages in months, and weights
in kilograms.
The length-for-age and stature-for-age methods were needed for the
blood pressure methods above.
All methods use the published LMS parameters to define z-scores,
percentiles, and quantiles for skewed distributions. L is a \(\lambda\) parameter, the Box-Cox
transformation power; \(M\) the median
value, and \(S\) a generalized
coefficient of variation. For a given percentile or z-score, the
corresponding physical measurement, \(X,\) is defined as
\[X = \begin{cases}
M \left(1 + \lambda S Z \right)^{\frac{1}{\lambda}} & \lambda
\neq 0 \\
M \exp\left( S Z \right) & \lambda = 0.
\end{cases}\]
From this we can get the z-score for a given measurement \(X:\)
\[ Z = \begin{cases}
\frac{\left(\frac{X}{M}\right)^{\lambda} - 1}{\lambda S} & \lambda
\neq 0 \\
\frac{\log\left(\frac{X}{M}\right) }{ S } & \lambda = 0.
\end{cases}\]
Percentiles are determined using the standard normal distribution of
z-scores.
For all eight of the noted methods we provide a distribution
function, quantile function, and function that returns z-scores.
Estimates for finer differences in age, for example, are possible for
these methods than the blood pressure methods. This is due to the
permissible linear interpolation of the LMS parameters for the CDC
charts whereas the blood pressure assessment is restricted to values
within a look up table.
Length and Stature
For Age
A 13 year old male standing 154 cm tall is in the 39.51th
percentile:
p_stature_for_age(q = 154, age = 13 * 12, male = 1L)
## [1] 0.3951253
To find the height corresponding to the 50th, 60th, and 75th
percentiles for height for 9.5-year old girls:
q_stature_for_age(p = c(0.50, 0.60, 0.75), age = 9.5 * 12, male = 0L)
## [1] 135.4252 137.0665 139.8316
If you want the standard score for a percentile, you can use qnorm
around p_stature_for_age, or simply call z_stature_for_age.
qnorm(p_stature_for_age(q = 154, age = 13 * 12, male = 1L))
## [1] -0.2659852
z_stature_for_age(q = 154, age = 13 * 12, male = 1L)
## [1] -0.2659852
A length/height for age chart based on the CDC data:
Weight for Age
There are two methods for determining weight for age both based on
CDC National Center for Health Statistics data: one for infants (weighed
laying flat) up to 36 months (“Growth
Charts - Data Table of Infant Weight-for-Age Charts”
2001), and one for children (weighed on a standing scale) over 24
months (“Growth Charts - Data Table of
Weight-for-Age Charts” 2001).
A 33 pound ( 14.97 kg) 4 year old male is in the 24.11th
percentile.
p_weight_for_age(33 * 0.453592, age = 4 * 12, male = 1)
## [1] 0.2410771
The 20th percentile weight for an 18 month old infant female is
10.07575 kg.
round(q_weight_for_age_inf(p = 0.2, age = 18, male = 0), 5)
## [1] 10.07575
Weight for Length or
Stature
Similar to weight-for-age, there are two methods for determining
weight-for-length. Both methods utilize data from the CDC National
Center for Health Statistics. The first method is used for infants under
36 months (used for children who are measured laying flat), and the
second is used for child over 24 months (used for children able to be
measured while standing up). The overlapping range between the methods
will differ.
The median weight for a 95 cm long infant 14.2224238 kg, whereas the
median weight for a 95 cm tall child is 14.4183484 kg.
q_weight_for_length_inf(0.5, 95, 1)
## [1] 14.22242
q_weight_for_stature(0.5, 95, 1)
## [1] 14.41835
A 5.8 kg, 61 cm long female infant is in the 0.2977671 weight
percentile.
p_weight_for_length_inf(5.8, 61, 0)
## [1] 0.2977671
BMI for Age
For a twelve year old, a BMI of 22.2 corresponds to the 87.28th BMI
percentile for a female, and the 90.36th BMI percentile for a male.
p_bmi_for_age(q = 22.2, age = c(144, 144), male = c(0, 1))
## [1] 0.8727830 0.9036248
The median BMI values for a 10 year old male and females are:
q_bmi_for_age(p = 0.5, age = c(120, 120), male = c(1, 0))
## [1] 16.62461 16.83800
Head
Circumference
A 10 month old male has a median head circumference of 45.6670663
cm.
A head circumference of 42 cm for an 8 month old female is in the
11.29397th percentile.
q_head_circ_for_age(0.5, 10, 1)
## [1] 45.66707
p_head_circ_for_age(42, 8, 0)
## [1] 0.1129397
Additional
Utilities
Estimating Gaussian
Mean and Standard Deviation
The NHLBI data for blood pressures provided values in percentiles. To
get a mean and standard deviation that would work well for estimating
other percentiles and quantiles via a Gaussian distribution we optimized
for values of the mean and standard deviation such that for the provided
quantiles \(q_i\) at the \(p_i\) percentiles and \(X \sim N\left(\mu, \sigma\right)\),
\[ \sum_{i} \left(\Pr(X \leq q_i) - p_i
\right)^2, \]
was minimized. The NHLBI data is provided to the end user.
data(list = "nhlbi_bp_norms", package = "pedbp")
str(nhlbi_bp_norms)
## 'data.frame': 952 obs. of 6 variables:
## $ male : int 0 0 0 0 0 0 0 0 0 0 ...
## $ age : num 12 12 12 12 12 12 12 12 12 12 ...
## $ height_percentile: int 5 5 5 5 10 10 10 10 25 25 ...
## $ bp_percentile : int 50 90 95 99 50 90 95 99 50 90 ...
## $ sbp : int 83 97 100 108 84 97 101 108 85 98 ...
## $ dbp : int 38 52 56 64 39 53 57 64 39 53 ...
For an example of how we fitted the parameters:
d <- nhlbi_bp_norms[nhlbi_bp_norms$age == 144 & nhlbi_bp_norms$height_percentile == 50, ]
d <- d[d$male == 0, ]
d
## male age height_percentile bp_percentile sbp dbp
## 321 0 144 50 50 105 62
## 322 0 144 50 90 119 76
## 323 0 144 50 95 123 80
## 324 0 144 50 99 130 88
est_norm(q = d$sbp, p = d$bp_percentile / 100)
## mean sd
## 105.00091 10.92092
est_norm(q = d$dbp, p = d$bp_percentile / 100)
## mean sd
## 61.99821 10.94227
bp_parameters[bp_parameters$male == 0 & bp_parameters$age == 144 & bp_parameters$height_percentile == 50, ]
## source male age sbp_mean sbp_sd dbp_mean dbp_sd height_percentile
## 89 nhlbi 0 144 105.0009 10.92092 61.99821 10.94227 50
## NA <NA> NA NA NA NA NA NA NA
The est_norm method comes with a plotting method too. The provided
quantiles are plotted as open dots and the fitted distribution function
is plotted to show the fit.
plot( est_norm(q = d$dbp, p = d$bp_percentile / 100) )
If you want to emphasize a data point you can do that as well. Here
is an example from a set of quantiles and percentiles which are not
Gaussian.
qs <- c(-1.92, 0.05, 0.1, 1.89) * 1.8 + 3.14
ps <- c(0.025, 0.40, 0.50, 0.975)
# with equal weights
w0 <- est_norm(qs, ps)
# weight to ignore one of the middle value and make sure to hit the other
w1 <- est_norm(qs, ps, weights = c(1, 2, 0, 1))
# equal weight the middle, more than the tails
w2 <- est_norm(qs, ps, weights = c(1, 2, 2, 1))
gridExtra::grid.arrange(
plot(w0) + ggplot2::ggtitle(label = "w0", subtitle = paste0("Mean: ", round(w0$par[1], 2), " SD: ", round(w0$par[2], 3)))
, plot(w1) + ggplot2::ggtitle(label = "w1", subtitle = paste0("Mean: ", round(w1$par[1], 2), " SD: ", round(w1$par[2], 3)))
, plot(w2) + ggplot2::ggtitle(label = "w2", subtitle = paste0("Mean: ", round(w2$par[1], 2), " SD: ", round(w2$par[2], 3)))
, nrow = 1
)
