我想把这个公式输入R:
该公式采用以下输入:
所有上述输入矢量18个元素长,因为它们指的是18个年龄间隔.前17个年龄间隔为5年,最后一个间隔(85年以上)无限宽.
该公式估计了Sasieni等人2011年提出的终身癌症风险 http://www.nature.com/bjc/journal/v105/n3/full/bjc2011250a.html
# Input data:
M <- c(140L, 12L, 12L, 59L, 94L, 101L, 117L, 213L, 368L, 607L, 1025L,
1488L, 2255L, 2787L, 3257L, 3715L, 4231L, 6281L)
R <- c(42L, 22L, 28L, 54L, 77L, 108L, 169L, 227L, 293L, 531L, 863L,
1464L, 2591L, 3334L, 3045L, 2605L, 1890L, 1261L)
D <- c(2L, 1L, 2L, 6L, 4L, 7L, 15L, 26L, 67L, 120L, 304L, 497L, 883L,
1158L, 1321L, 1318L, 1177L, 1065L)
N <- c(167323L, 168088L, 176017L, 180986L, 168189L, 155506L, 174274L,
195538L, 207287L, 204711L, 183802L, 174342L, 183415L, 151277L,
104199L, 71782L, 47503L, 33946L)
# W width of age interval
w <- c( 5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,Inf )
# function
v1 <- numeric()
for(i in 1:length(R)) {
v1[i] <- R[i] / ( R[i] + M[i] - D[i] ) * ( 1 - exp( - (w[i]/N[i]) * (R[i] + M[i] - D[i]) ) )
}
sum(v1)
Run Code Online (Sandbox Code Playgroud)
代码看起来像代码一样尽可能多的答案是首选,因此不了解R的同事可以识别代码中的等式.
答案应该是0.376127241057822
raw*_*awr 12
也许这会奏效.你可以检查一下纸上的例子吗?
f <- function(idx) {
s <- numeric(idx)
for (i in 1:idx)
s[i] <- R[i] / (R[i] + M[i] - D[i]) * S(i) * (1 - exp(-w[i] / N[i] * (R[i] + M[i] - D[i])))
s
}
S <- function(idx) {
if (idx == 1L)
return(1)
s <- numeric(idx - 1)
for (j in 1:(idx - 1))
s[j] <- (R[j] + (M[j] - D[j])) / N[j]
exp(-sum(s))
}
# Input data:
M <- c(140L, 12L, 12L, 59L, 94L, 101L, 117L, 213L, 368L, 607L, 1025L,
1488L, 2255L, 2787L, 3257L, 3715L, 4231L, 6281L)
R <- c(42L, 22L, 28L, 54L, 77L, 108L, 169L, 227L, 293L, 531L, 863L,
1464L, 2591L, 3334L, 3045L, 2605L, 1890L, 1261L)
D <- c(2L, 1L, 2L, 6L, 4L, 7L, 15L, 26L, 67L, 120L, 304L, 497L, 883L,
1158L, 1321L, 1318L, 1177L, 1065L)
N <- c(167323L, 168088L, 176017L, 180986L, 168189L, 155506L, 174274L,
195538L, 207287L, 204711L, 183802L, 174342L, 183415L, 151277L,
104199L, 71782L, 47503L, 33946L)
# W width of age interval
w <- c( 5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,Inf )
f(18)
# [1] 0.0012516883 0.0006533947 0.0007939380 0.0014874104 0.0022786758 0.0034506651
# [7] 0.0048088199 0.0057397672 0.0069608906 0.0126706127 0.0226156951 0.0395612334
# [13] 0.0644167605 0.0956951717 0.1184236481 0.1330917708 0.1256574840 0.1421444626
sum(f(18))
# [1] 0.7817021
Run Code Online (Sandbox Code Playgroud)
一种更"R"的方式
lr <- length(R)
S <- sapply(seq(R), function(idx)
exp(-sum((R[-(idx:lr)] + (M[-(idx:lr)] - D[-(idx:lr)])) / N[-(idx:lr)])))
sum(R / (R + M - D) * S * (1 - exp(-w / N * (R + M - D))))
# [1] 0.7817021
Run Code Online (Sandbox Code Playgroud)
| 归档时间: |
|
| 查看次数: |
840 次 |
| 最近记录: |