有条件地替换 data.frame 中的值 [英] Conditional replacement of values in a data.frame

问题描述

我试图了解如何在不使用循环的情况下有条件地替换数据框中的值.我的数据框结构如下:

I am trying to understand how to conditional replace values in a dataframe without using a loop. My data frame is structured as follows:

> df
          a b est
1  11.77000 2   0
2  10.90000 3   0
3  10.32000 2   0
4  10.96000 0   0
5   9.90600 0   0
6  10.70000 0   0
7  11.43000 1   0
8  11.41000 2   0
9  10.48512 4   0
10 11.19000 0   0

和 dput 输出是这样的:

structure(list(a = c(11.77, 10.9, 10.32, 10.96, 9.906, 10.7, 
11.43, 11.41, 10.48512, 11.19), b = c(2, 3, 2, 0, 0, 0, 1, 2, 
4, 0), est = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0)), .Names = c("a", 
"b", "est"), row.names = c(NA, -10L), class = "data.frame")

我想做的是检查b的值.如果 b 为 0，我想将 est 设置为 a 中的值.我知道 df$est[df$b == 0] <- 23 会将 est 的所有值设置为 23，当 b==0.我不明白的是如何在该条件为真时将 est 设置为 a 的值.例如:

What I want to do, is to check the value of b. If b is 0, I want to set est to a value from a. I understand that df$est[df$b == 0] <- 23 will set all values of est to 23, when b==0. What I don't understand is how to set est to a value of a when that condition is true. For example:

df$est[df$b == 0] <- (df$a - 5)/2.533 

给出以下警告:

Warning message:
In df$est[df$b == 0] <- (df$a - 5)/2.533 :
  number of items to replace is not a multiple of replacement length

有没有办法可以传递相关单元格而不是向量?

Is there a way that I can pass the relevant cell, rather than vector?

推荐答案

既然你是有条件地索引 df$est，你还需要有条件地索引替换向量 df$a代码>:

Since you are conditionally indexing df$est, you also need to conditionally index the replacement vector df$a:

index <- df$b == 0
df$est[index] <- (df$a[index] - 5)/2.533 

当然，变量 index 只是临时的，我用它来使代码更易读.一步就可以写出来:

Of course, the variable index is just temporary, and I use it to make the code a bit more readible. You can write it in one step:

df$est[df$b == 0] <- (df$a[df$b == 0] - 5)/2.533 

为了更好的可读性，你可以使用 within:

For even better readibility, you can use within:

df <- within(df, est[b==0] <- (a[b==0]-5)/2.533)

结果，无论您选择哪种方法:

The results, regardless of which method you choose:

df
          a b      est
1  11.77000 2 0.000000
2  10.90000 3 0.000000
3  10.32000 2 0.000000
4  10.96000 0 2.352941
5   9.90600 0 1.936834
6  10.70000 0 2.250296
7  11.43000 1 0.000000
8  11.41000 2 0.000000
9  10.48512 4 0.000000
10 11.19000 0 2.443743

<小时>

正如其他人所指出的，您的示例中的替代解决方案是使用 ifelse.

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