R:滤波器问题 - `filter()`::!计算时问题`..1 =(数据$ column,400,1800)'

发布于 2025-01-26 04:52:09 字数 684 浏览 3 评论 0原文

我有一个数据框架,并希望根据与另一列有关的条件获得子集。这似乎适用于某个变量,但是对于一个特定的变量,我会收到以下错误消息:

Error in `filter()`:
! Problem while computing `..1 = between(Unitranche$Yield, 400, 1800)`.
Caused by error in `Unitranche$Yield`:
! $ operator is invalid for atomic vectors
Run `rlang::last_error()` to see where the error occurred.

代码如下:

X = filter(data, X == 1)
Y = filter(data, Y == 1)
Z = filter(data), Z == 1)

直到这里没有问题,但是:

X = filter(X, between(X$a,400,1500))
Y = filter(Y, between(Y$a,400,1800))
Z = filter(Z, a >= 800)

尽管y是创建/过滤y的,但该错误弹出了y。就像X和Z一样。

不确定如何使用示例数据复制此错误;但是,以防万一是5个变量的278个观察结果。

谢谢你!

I have a data frame and am looking to get a subset based on conditions related to another column. This seems to work for some variable, but somehow for one particular one, I am getting the following error message:

Error in `filter()`:
! Problem while computing `..1 = between(Unitranche$Yield, 400, 1800)`.
Caused by error in `Unitranche$Yield`:
! $ operator is invalid for atomic vectors
Run `rlang::last_error()` to see where the error occurred.

The code is as follows:

X = filter(data, X == 1)
Y = filter(data, Y == 1)
Z = filter(data), Z == 1)

no problem until here, but then:

X = filter(X, between(X$a,400,1500))
Y = filter(Y, between(Y$a,400,1800))
Z = filter(Z, a >= 800)

The error pops up for Y, although Y is created/filtered exactly like the X and Z.

Not sure how to replicate this error with sample data; but just in case Y is 278 observations of 5 variables.

Thank you!

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