内插 NA 值

发布于 2024-12-01 11:49:22 字数 820 浏览 1 评论 0原文

我有两组与时间无关的样本。我想合并它们并计算缺失值 在我不具备两者价值观的时代。简化示例:

A <- cbind(time=c(10, 20, 30, 40, 50, 60, 70, 80, 90, 100),
           Avalue=c(1, 2, 3, 2, 1, 2, 3, 2, 1, 2))
B <- cbind(time=c(15, 30, 45, 60), Bvalue=c(100, 200, 300, 400))
C <- merge(A,B, all=TRUE)

   time Avalue Bvalue
1    10      1     NA
2    15     NA    100
3    20      2     NA
4    30      3    200
5    40      2     NA
6    45     NA    300
7    50      1     NA
8    60      2    400
9    70      3     NA
10   80      2     NA
11   90      1     NA
12  100      2     NA

通过假设每个样本之间的线性变化,可以计算缺失的 NA 值。 直观上很容易看出时间15和45处的A值应该是1.5。但对 B 的正确计算 例如,时间 20 为

100 + (20 - 15) * (200 - 100) / (30 - 15)

,等于 133.33333。 第一个括号是估计时间和最后一个可用样本之间的时间。 第二个括号是最近样本之间的差异。 第三个括号是最近样本之间的时间。

如何使用 R 计算 NA 值?

I have two set of samples that are time independent. I would like to merge them and calculate the missing values
for the times where I do not have values of both. Simplified example:

A <- cbind(time=c(10, 20, 30, 40, 50, 60, 70, 80, 90, 100),
           Avalue=c(1, 2, 3, 2, 1, 2, 3, 2, 1, 2))
B <- cbind(time=c(15, 30, 45, 60), Bvalue=c(100, 200, 300, 400))
C <- merge(A,B, all=TRUE)

   time Avalue Bvalue
1    10      1     NA
2    15     NA    100
3    20      2     NA
4    30      3    200
5    40      2     NA
6    45     NA    300
7    50      1     NA
8    60      2    400
9    70      3     NA
10   80      2     NA
11   90      1     NA
12  100      2     NA

By assuming linear change between each sample, it is possible to calculate the missing NA values.
Intuitively it is easy to see that the A value at time 15 and 45 should be 1.5. But a proper calculation for B
for instance at time 20 would be

100 + (20 - 15) * (200 - 100) / (30 - 15)

which equals 133.33333.
The first parenthesis being the time between estimate time and the last sample available.
The second parenthesis being the difference between the nearest samples.
The third parenthesis being the time between the nearest samples.

How can I use R to calculate the NA values?

如果你对这篇内容有疑问,欢迎到本站社区发帖提问 参与讨论,获取更多帮助,或者扫码二维码加入 Web 技术交流群。

扫码二维码加入Web技术交流群

发布评论

需要 登录 才能够评论, 你可以免费 注册 一个本站的账号。

评论(3

贱贱哒 2024-12-08 11:49:22

使用 zoo 包:

library(zoo)
Cz <- zoo(C)
index(Cz) <- Cz[,1]
Cz_approx <- na.approx(Cz)

Using the zoo package:

library(zoo)
Cz <- zoo(C)
index(Cz) <- Cz[,1]
Cz_approx <- na.approx(Cz)
蓝礼 2024-12-08 11:49:22

进行统计分析并仍然获得有效置信区间的正确方法是使用多重插补。请参阅 Rubin 的经典,这是一本出色的 R 此包 (mi)

The proper way to do this statistically and still get valid confidence intervals is to use Multiple Imputation. See Rubin's classic book, and there's an excellent R package for this (mi).

惟欲睡 2024-12-08 11:49:22

一个丑陋且可能效率低下的 Base R 解决方案:

# Data provided:
A <- cbind(time=c(10, 20, 30, 40, 50, 60, 70, 80, 90, 100),
           Avalue=c(1, 2, 3, 2, 1, 2, 3, 2, 1, 2))
B <- cbind(time=c(15, 30, 45, 60), Bvalue=c(100, 200, 300, 400))
C <- merge(A,B, all=TRUE)

# Scalar valued at the minimum time difference: -> min_time_diff

min_time_diff <- min(diff(C$time))

# Adjust frequency of the series to hold all steps in range: -> df

df <- merge(C, 
            data.frame(time = seq(min_time_diff, 
                                 max(C$time), 
                                 by = min_time_diff)),
           by = "time",
           all = TRUE)



# Linear interpolation function handling ties,
# returns interpolated vector the same length 
# a the input vector: -> vector

l_interp_vec <- function(na_vec){

  approx(x = na_vec,

         method = "linear",

         ties = "constant",

         n = length(na_vec))$y

}

# Applied to a dataframe, replacing NA values
# in each of the numeric vectors, 
# with interpolated values. 
# input is dataframe: -> dataframe()

interped_df <- data.frame(lapply(df, function(x){

      if(is.numeric(x)){

        # Store a scalar of min row where x isn't NA: -> min_non_na

        min_non_na <- min(which(!(is.na(x))))

        # Store a scalar of max row where x isn't NA: -> max_non_na

        max_non_na <- max(which(!(is.na(x))))

        # Store scalar of the number of rows needed to impute prior 
        # to first NA value: -> ru_lower

        ru_lower <- ifelse(min_non_na > 1, min_non_na - 1, min_non_na)

        # Store scalar of the number of rows needed to impute after
        # the last non-NA value: -> ru_lower

        ru_upper <- ifelse(max_non_na == length(x), 

                           length(x) - 1, 

                           (length(x) - (max_non_na + 1)))

        # Store a vector of the ramp to function: -> l_ramp_up: 

        ramp_up <- as.numeric(
          cumsum(rep(x[min_non_na]/(min_non_na), ru_lower))
          )

        # Apply the interpolation function on vector "x": -> y

        y <- as.numeric(l_interp_vec(as.numeric(x[min_non_na:max_non_na])))

        # Create a vector that combines the ramp_up vector 
        # and y if the first NA is at row 1: -> z

        if(length(ramp_up) > 1 & max_non_na != length(x)){

          # Create a vector interpolations if there are 
          # multiple NA values after the last value: -> lower_l_int

          lower_l_int <- as.numeric(cumsum(rep(mean(diff(c(ramp_up, y))),
                                               ru_upper+1)) +
                                  as.numeric(x[max_non_na]))

          # Store the linear interpolations in  a vector: -> z

          z <- as.numeric(c(ramp_up, y, lower_l_int))

        }else if(length(ramp_up) > 1 & max_non_na == length(x)){

          # Store the linear interpolations in  a vector: -> z

          z <- as.numeric(c(ramp_up, y))

        }else if(min_non_na == 1 & max_non_na != length(x)){

          # Create a vector interpolations if there are 
          # multiple NA values after the last value: -> lower_l_int

          lower_l_int <- as.numeric(cumsum(rep(mean(diff(c(ramp_up, y))),
                                               ru_upper+1)) +
                                  as.numeric(x[max_non_na]))


          # Store the linear interpolations in  a vector: -> z

          z <- as.numeric(c(y, lower_l_int))

        }else{

          # Store the linear interpolations in  a vector: -> z

          z <- as.numeric(y)

        }

        # Interpolate between points in x, return new x:

        return(as.numeric(ifelse(is.na(x), z, x)))

      }else{

        x

      }

    }

  )

)

# Subset interped df to only contain 
# the time values in C, store a data frame: -> int_df_subset

int_df_subset <- interped_df[interped_df$time %in% C$time,]

An ugly and probably inefficient Base R solution:

# Data provided:
A <- cbind(time=c(10, 20, 30, 40, 50, 60, 70, 80, 90, 100),
           Avalue=c(1, 2, 3, 2, 1, 2, 3, 2, 1, 2))
B <- cbind(time=c(15, 30, 45, 60), Bvalue=c(100, 200, 300, 400))
C <- merge(A,B, all=TRUE)

# Scalar valued at the minimum time difference: -> min_time_diff

min_time_diff <- min(diff(C$time))

# Adjust frequency of the series to hold all steps in range: -> df

df <- merge(C, 
            data.frame(time = seq(min_time_diff, 
                                 max(C$time), 
                                 by = min_time_diff)),
           by = "time",
           all = TRUE)



# Linear interpolation function handling ties,
# returns interpolated vector the same length 
# a the input vector: -> vector

l_interp_vec <- function(na_vec){

  approx(x = na_vec,

         method = "linear",

         ties = "constant",

         n = length(na_vec))$y

}

# Applied to a dataframe, replacing NA values
# in each of the numeric vectors, 
# with interpolated values. 
# input is dataframe: -> dataframe()

interped_df <- data.frame(lapply(df, function(x){

      if(is.numeric(x)){

        # Store a scalar of min row where x isn't NA: -> min_non_na

        min_non_na <- min(which(!(is.na(x))))

        # Store a scalar of max row where x isn't NA: -> max_non_na

        max_non_na <- max(which(!(is.na(x))))

        # Store scalar of the number of rows needed to impute prior 
        # to first NA value: -> ru_lower

        ru_lower <- ifelse(min_non_na > 1, min_non_na - 1, min_non_na)

        # Store scalar of the number of rows needed to impute after
        # the last non-NA value: -> ru_lower

        ru_upper <- ifelse(max_non_na == length(x), 

                           length(x) - 1, 

                           (length(x) - (max_non_na + 1)))

        # Store a vector of the ramp to function: -> l_ramp_up: 

        ramp_up <- as.numeric(
          cumsum(rep(x[min_non_na]/(min_non_na), ru_lower))
          )

        # Apply the interpolation function on vector "x": -> y

        y <- as.numeric(l_interp_vec(as.numeric(x[min_non_na:max_non_na])))

        # Create a vector that combines the ramp_up vector 
        # and y if the first NA is at row 1: -> z

        if(length(ramp_up) > 1 & max_non_na != length(x)){

          # Create a vector interpolations if there are 
          # multiple NA values after the last value: -> lower_l_int

          lower_l_int <- as.numeric(cumsum(rep(mean(diff(c(ramp_up, y))),
                                               ru_upper+1)) +
                                  as.numeric(x[max_non_na]))

          # Store the linear interpolations in  a vector: -> z

          z <- as.numeric(c(ramp_up, y, lower_l_int))

        }else if(length(ramp_up) > 1 & max_non_na == length(x)){

          # Store the linear interpolations in  a vector: -> z

          z <- as.numeric(c(ramp_up, y))

        }else if(min_non_na == 1 & max_non_na != length(x)){

          # Create a vector interpolations if there are 
          # multiple NA values after the last value: -> lower_l_int

          lower_l_int <- as.numeric(cumsum(rep(mean(diff(c(ramp_up, y))),
                                               ru_upper+1)) +
                                  as.numeric(x[max_non_na]))


          # Store the linear interpolations in  a vector: -> z

          z <- as.numeric(c(y, lower_l_int))

        }else{

          # Store the linear interpolations in  a vector: -> z

          z <- as.numeric(y)

        }

        # Interpolate between points in x, return new x:

        return(as.numeric(ifelse(is.na(x), z, x)))

      }else{

        x

      }

    }

  )

)

# Subset interped df to only contain 
# the time values in C, store a data frame: -> int_df_subset

int_df_subset <- interped_df[interped_df$time %in% C$time,]
~没有更多了~
我们使用 Cookies 和其他技术来定制您的体验包括您的登录状态等。通过阅读我们的 隐私政策 了解更多相关信息。 单击 接受 或继续使用网站,即表示您同意使用 Cookies 和您的相关数据。
原文