我在dplyr中遇到mutate函数时遇到问题，错误说明了
Error: incompatible size (0), expecting 5 (the group size) or 1
之前有一些帖子，我尝试了一些解决方案，但我的情况没有运气。
group-factorial-data-with-multiple-factors-error-incompatible-size-0-expe
r-dplyr-using-mutate-with-na-omit-causes-error-incompatible-size-d
grouped-operations-that-result-in-length-not-equal-to-1-or-length-of-group-in-dp
这是我尝试过的，
ff <- c(seq(0,0.2,0.1),seq(0,-0.2,-0.1))
flip <- c(c(0,0,1,1,1,1),c(1,1,0,0,0,0))
df <- data.frame(ff,flip,group=gl(2,6))
> df
ff flip group
1 0.0 0 1
2 0.1 0 1
3 0.2 1 1
4 0.0 1 1
5 -0.1 1 1
6 -0.2 1 1
7 0.0 1 2
8 0.1 1 2
9 0.2 0 2
10 0.0 0 2
11 -0.1 0 2
12 -0.2 0 2
我想根据以下某些条件添加名为c1和c2的新群组
dff <- df%>%
group_by(group)%>%
mutate(flip=as.numeric(flip),direc=ifelse(c(0,diff(ff))<0,"backward","forward"))%>%
spread(direc,flip)%>%
arrange(group,group)%>%
mutate(c1=ff[head(which(forward>0),1)],c2=ff[tail(which(backward>0),1)])
错误：大小不一致（0），期望5（组大小）或1
我还添加了do并尝试了
do(data.frame(., c1=ff[head(which(.$forward>0),1)],c2=ff[tail(which(.$backward>0),1)]))
data.frame中的错误（。，c1 = ff [head（其中。$ forward＆gt; 0），1）]，c2 = ff [tail（其中。$ $＆gt;： 参数意味着不同的行数：5,1,0
但是当我只有mutate c1列时，一切似乎都在起作用。为什么呢？
答案 0 :(得分：3)
通过管道查看发生的情况可能会提供信息。
df %>%
group_by(group)%>%
mutate(flip=as.numeric(flip),direc=ifelse(c(0,diff(ff))<0,"backward","forward"))%>%
spread(direc,flip)%>%
arrange(group,group)
# Source: local data frame [10 x 4]
# Groups: group [2]
# ff group backward forward
# <dbl> <fctr> <dbl> <dbl>
# 1 -0.2 1 1 NA
# 2 -0.1 1 1 NA
# 3 0.0 1 1 0
# 4 0.1 1 NA 0
# 5 0.2 1 NA 1
# 6 -0.2 2 0 NA
# 7 -0.1 2 0 NA
# 8 0.0 2 0 1
# 9 0.1 2 NA 1
# 10 0.2 2 NA 0
BTW：为什么arrange(group,group)？加倍订单变量毫无意义。
在此处，您会看到您有（1）backward个值不大于0.当您运行which(FALSE)之类的内容时，您会获得integer(0)。这可能是认识到dplyr需要rhs的向量长度与组中行数相同的长度的好时机。
而不是mutate，我会稍微修改一下：返回which c2调用中返回的唯一值的数量：
df %>%
group_by(group)%>%
mutate(flip=as.numeric(flip),direc=ifelse(c(0,diff(ff))<0,"backward","forward"))%>%
spread(direc,flip)%>%
arrange(group,group)%>%
mutate(
c1 = ff[head(which(forward>0),1)],
c2len = length(which(backward > 0))
)
# Source: local data frame [10 x 6]
# Groups: group [2]
# ff group backward forward c1 c2len
# <dbl> <fctr> <dbl> <dbl> <dbl> <int>
# 1 -0.2 1 1 NA 0.2 3
# 2 -0.1 1 1 NA 0.2 3
# 3 0.0 1 1 0 0.2 3
# 4 0.1 1 NA 0 0.2 3
# 5 0.2 1 NA 1 0.2 3
# 6 -0.2 2 0 NA 0.0 0
# 7 -0.1 2 0 NA 0.0 0
# 8 0.0 2 0 1 0.0 0
# 9 0.1 2 NA 1 0.0 0
# 10 0.2 2 NA 0 0.0 0
为了在ff上进行有意义的索引，您需要在退货中使用integer(0)以外的其他内容。
答案 1 :(得分：3)
只是扩展@ allistaire的评论。
tail(which(backward>0),1) spread() 你可以尝试
dff <- df%>%
group_by(group)%>%
mutate(flip=as.numeric(flip),direc=ifelse(c(0,diff(ff))<0,"backward","forward"))%>%
arrange(group)%>%
mutate(c1=ff[head(which(direc=="forward" & flip > 0),1)])
对于每个direction，您似乎希望确定group更改的涌入点。在这种情况下，请详细说明翻转是如何相关的，或者如果您将flip <- c(c(0,0,1,1,1,1),c(1,1,0,0,0,0))更改为flip <- c(c(0,0,1,1,1,1),c(1,1,0,1,1,1))以便flip标记ff的方向发生变化，则可以使用< / p>
dff <- df%>%
group_by(group)%>%
mutate(flip=as.numeric(flip),direc=ifelse(c(0,diff(ff))<0,"backward","forward"))%>%
arrange(group)%>%
mutate(c1=ff[head(which(direc=="forward" & flip > 0),1)]) %>%
mutate(c2=ff[tail(which(direc=="backward"& flip >0),1)])
给出：
Source: local data frame [12 x 6]
Groups: group [2]
ff flip group direc c1 c2
<dbl> <dbl> <fctr> <chr> <dbl> <dbl>
1 0.0 0 1 forward 0.2 -0.2
2 0.1 0 1 forward 0.2 -0.2
3 0.2 1 1 forward 0.2 -0.2
4 0.0 1 1 backward 0.2 -0.2
5 -0.1 1 1 backward 0.2 -0.2
6 -0.2 1 1 backward 0.2 -0.2
7 0.0 1 2 forward 0.0 -0.2
8 0.1 1 2 forward 0.0 -0.2
9 0.2 0 2 forward 0.0 -0.2
10 0.0 1 2 backward 0.0 -0.2
11 -0.1 1 2 backward 0.0 -0.2
12 -0.2 1 2 backward 0.0 -0.2