在R中使用线性回归填充NA [英] Filling NA using linear regression in R

问题描述

我有一个带有一个时间列和2个变量的数据.(下面的示例)

I have a data with one time column and 2 variables.(example below)

df <- structure(list(time = c(15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 
                              25, 26), var1 = c(20.4, 31.5, NA, 53.7, 64.8, NA, NA, NA, NA, 
                              120.3, NA, 142.5), var2 = c(30.6, 47.25, 63.9, 80.55, 97.2, 113.85, 
                              130.5, 147.15, 163.8, 180.45, 197.1, 213.75)), .Names = c("time", 
                              "var1", "var2"), row.names = c(NA, -12L), class = c("tbl_df", 
                              "tbl", "data.frame"))

var1 的 NA 很少，我想用 var1 和 var2 中剩余值之间的线性回归填充 NA.

The var1 has few NA and I want to fill the NA with linear regression between remaining values in var1 and var2.

请帮助！！如果您需要更多信息，请告诉我

Please Help!! And let me know if you need more information

推荐答案

以下是使用 lm 预测R中值的示例.

Here is an example using lm to predict values in R.

library(dplyr)

# Construct linear model based on non-NA pairs
df2 <- df %>% filter(!is.na(var1))

fit <- lm(var1 ~ var2, data = df2)

# See the result
summary(fit)

# Call:
#   lm(formula = var1 ~ var2, data = df2)
# 
# Residuals:
#   1          2          3          4          5          6 
# 8.627e-15 -2.388e-15  1.546e-16 -9.658e-15 -2.322e-15  5.587e-15 
# 
# Coefficients:
#   Estimate Std. Error   t value Pr(>|t|)    
# (Intercept) 2.321e-14  5.619e-15 4.130e+00   0.0145 *  
#   var2        6.667e-01  4.411e-17 1.511e+16   <2e-16 ***
#   ---
#   Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
# 
# Residual standard error: 7.246e-15 on 4 degrees of freedom
# Multiple R-squared:      1,   Adjusted R-squared:      1 
# F-statistic: 2.284e+32 on 1 and 4 DF,  p-value: < 2.2e-16
# 
# Warning message:
#   In summary.lm(fit) : essentially perfect fit: summary may be unreliable

# Use fit to predict the value
df3 <- df %>% 
  mutate(pred = predict(fit, .)) %>%
  # Replace NA with pred in var1
  mutate(var1 = ifelse(is.na(var1), pred, var1))

# See the result
df3 %>% as.data.frame()

#    time  var1   var2  pred
# 1    15  20.4  30.60  20.4
# 2    16  31.5  47.25  31.5
# 3    17  42.6  63.90  42.6
# 4    18  53.7  80.55  53.7
# 5    19  64.8  97.20  64.8
# 6    20  75.9 113.85  75.9
# 7    21  87.0 130.50  87.0
# 8    22  98.1 147.15  98.1
# 9    23 109.2 163.80 109.2
# 10   24 120.3 180.45 120.3
# 11   25 131.4 197.10 131.4
# 12   26 142.5 213.75 142.5

这篇关于在R中使用线性回归填充NA的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！