估算缺失数据,同时强制相关系数保持不变

Question

考虑以下(excel)数据集:

m   |   r
----|------
2.0 | 3.3
0.8 |   
    | 4.0
1.3 |   
2.1 | 5.2
    | 2.3
    | 1.9
2.5 | 
1.2 | 3.0
2.0 | 2.6

我的目标是使用以下条件填写缺失值:

将上述两列(约0.68)之间的成对相关性表示为R. 在填充空单元格之后将相关性表示为R*.填写表格以使(R-R*)^ 2 = 0.这是,我想保持数据的相关结构完整.

到目前为止,我已经使用Matlab完成了它:

clear all;

m = xlsread('data.xlsx','A2:A11') ;
r = xlsread('data.xlsx','B2:B11') ;

rho = corr(m,r,'rows','pairwise');

x0 = [1,1,1,1,1,1];
lb = [0,0,0,0,0,0];
f = @(x)my_correl(x,rho);

SOL = fmincon(f,x0,[],[],[],[],lb)

功能my_correl是:

function X = my_correl(x,rho)

sum_m = (11.9 + x(1) + x(2) + x(3));
sum_r = (22.3 + x(1) + x(2) + x(3));
avg_m = (11.9 + x(1) + x(2) + x(3))/8;
avg_r = (22.3 + x(4) + x(5) + x(6))/8;
rho_num = 8*(26.32 + 4*x(1) + 2.3*x(2) + 1.9*x(3) + 0.8*x(4) + 1.3*x(5) + 2.5*x(6)) - sum_m*sum_r;
rho_den = sqrt(8*(22.43 + (4*x(1))^2 + (2.3*x(2))^2 + (1.9*x(3))^2) - sum_m^2)*sqrt(8*(78.6 + (0.8*x(4))^2 + (1.3*x(5))^ + (2.5*x(6))^2) - sum_r^2);

X = (rho - rho_num/rho_den)^2;

end

此函数手动计算相关性,其中每个缺失的数据都是变量x(i).

问题是:我的实际数据集有超过20,000个观测值.我无法手动创建rho公式.

我怎样才能填写我的数据集？

注1:我愿意使用Python,Julia或R. Matlab等替代语言,这只是我的默认语言.

注2:答案将获得100分的奖励.从现在起承诺.

Answer 1

这是我接近它的方式,在R中提供了一个实现:

没有用于输入缺失数据点的唯一解决方案,使得完整(插补)数据的成对相关等于不完整数据的成对相关.因此,为了找到"好"解决方案而不仅仅是"任何"解决方案,我们可以引入一个额外的标准,即完整的估算数据也应该与原始数据共享相同的线性回归.这导致我们采用一种相当简单的方法.

1. 计算原始数据的线性回归模型.
2. 找到错误值的估算值,这些值将完全位于此回归线上.
3. 为该回归线周围的插补值生成残差的随机散布
4. 缩放估算的残差以迫使完整的估算数据的相关性等于原始数据的相关性

R中这样的解决方案:

library(data.table)
set.seed(123)

rho = cor(dt$m,dt$r,'pairwise')

# calculate linear regression of original data
fit1 = lm(r ~ m, data=dt)
fit2 = lm(m ~ r, data=dt)
# extract the standard errors of regression intercept (in each m & r direction)
# and multiply s.e. by sqrt(n) to get standard deviation 
sd1 = summary(fit1)$coefficients[1,2] * sqrt(dt[!is.na(r), .N])
sd2 = summary(fit2)$coefficients[1,2] * sqrt(dt[!is.na(m), .N])

# find where data points with missing values lie on the regression line
dt[is.na(r), r.imp := coefficients(fit1)[1] + coefficients(fit1)[2] * m] 
dt[is.na(m), m.imp := coefficients(fit2)[1] + coefficients(fit2)[2] * r]

# generate randomised residuals for the missing data, using the s.d. calculated above
dt[is.na(r), r.ran := rnorm(.N, sd=sd1)] 
dt[is.na(m), m.ran := rnorm(.N, sd=sd2)] 

# function that scales the residuals by a factor x, then calculates how close correlation of imputed data is to that of original data
obj = function(x, dt, rho) {
  dt[, r.comp := r][, m.comp := m]
  dt[is.na(r), r.comp := r.imp + r.ran*x] 
  dt[is.na(m), m.comp := m.imp + m.ran*x] 
  rho2 = cor(dt$m.comp, dt$r.comp,'pairwise')
  (rho-rho2)^2
}

# find the value of x that minimises the discrepencay of imputed versus original correlation
fit = optimize(obj, c(-5,5), dt, rho)

x=fit$minimum
dt[, r.comp := r][, m.comp := m]
dt[is.na(r), r.comp := r.imp + r.ran*x] 
dt[is.na(m), m.comp := m.imp + m.ran*x] 
rho2 = cor(dt$m.comp, dt$r.comp,'pairwise')
(rho-rho2)^2  # check that rho2 is approximately equal to rho

作为最终检查,计算完整的插补数据的线性回归并绘制以显示回归线与原始数据相同.请注意,下图是针对下面显示的大数据集,以演示此方法在大数据上的使用.

fit.comp = lm(r.comp ~ m.comp, data=dt)
plot(dt$m.comp, dt$r.comp)
points(dt$m, dt$r, col="red")
abline(fit1, col="green")
abline(fit.comp, col="blue")
mtext(paste(" Rho =", round(rho,5)), at=-1)
mtext(paste(" Rho2 =", round(rho2, 5)), at=6)

数据

OP示例中的原始玩具数据:

dt=structure(list(m = c(2, 0.8, NA, 1.3, 2.1, NA, NA, 2.5, 1.2, 2), 
                  r = c(3.3, NA, 4, NA, 5.2, 2.3, 1.9, NA, 3, 2.6)), 
             .Names = c("m", "r"), row.names = c(NA, -10L), 
             class = c("data.table", "data.frame"))

用于演示大数据的更大数据集

dt = data.table(m=rnorm(1e5, 3, 2))[, r:=1.5 + 1.1*m + rnorm(1e5,0,2)]
dt[sample(.N, 3e4), m:=NA]
dt[sample(which(!is.na(m)), 3e4), r := NA]