循环遍历R中后续行中的NA值 [英] Loop over NA values in subsequent rows in R

问题描述

我想用上一个乘以标量(例如3)替换数据框第一列中的每个缺失值

I want to replace each missing value in the first column of my dataframe with the previous one multiplied by a scalar (eg. 3)

nRowsDf <- nrow(df)
for(i in 1:nRowsDf){
  df[i,1] =ifelse(is.na(df[i,1]), lag(df[i,1])+3*lag(df[i,1]),   df[i,1])
  }

上面的代码没有给我一个错误，但也没有执行任何操作.

The above code does not give me an error but does not do the job either.

此外，有没有比编写循环更好的方法了?

In addition, is there a better way to do this instead of writing a loop?

更新和数据:

这里是数据示例.我想用上一个乘以标量(例如3)替换数据框第一列中的每个缺失值. NA值在随后的行中.

Here is an example of data. I want to replace each missing value in the first column of my dataframe with the previous one multiplied by a scalar (eg. 3). The NA values are in subsequent rows.

df <- mtcars
df[c(2,3,4,5),1] <-NA

IND <- is.na(df[,1])
df[IND,1] <- df[dplyr::lead(IND,1L, F),1] * 3

上面代码的最后一行逐行完成作业(我应该运行它4次以填充缺少的4行).如何为所有行一次?

The last line of the above code does the job row by row (I should run it 4 times to fill the 4 missing rows). How can I do it once for all rows?

推荐答案

您应提供的可复制数据:

reproducible data which YOU should provide:

df <- mtcars
df[c(1,5,8),1] <-NA

代码:

IND <- is.na(df[,1])
df[IND,1] <- df[dplyr::lag(IND,1L, F),1] * 3

---

• 因为您使用了滞后，所以我使用了滞后.您说的是上一个".因此，也许您想使用lead.
• 如果缺少前导情况下的第一个值或滞后情况下的最后一个值，会发生什么情况. (这仍然是个谜)

---

• since you use lag I use lag. You are saying "previous". So maybe you want to use lead.
• What happens if the first value in lead case or last value in lag case is missing. (this remains a mystery)

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