r - 在x首先超过y的组中过滤

我有一个数据库，由

家庭社区的

社区ID(id_h)，

Home块的

块ID(blk_h，

邻域的子地理)，

工作块(blk_w)，

两者之间的通勤流量(Flow)，

每个家庭社区的通勤者中位数(Med_C)和

按居委会(CumFlow)累积的工作人员流。

数据按blk_h和blk_w之间的距离(降序)排序，并按id_h分组。我需要对数据进行子集化，以提取CumFlow FIRST等于或超过Med_C的每个家庭邻居的情况。

我尝试了各种dplyr函数，但无法使其正常工作。这是一个例子:

df <- data.frame(
  id_h=c("A","A","A","A","B","B","B"),
  blk_h=c("A1","A1","A2","A2","B1","B2","B2"),
  blk_w=c("W1","W2","W3","W3","W1","W2","W2"),
  dist=c(4.3,5.6,7.0,8.7,5.2,6.5,6.8),
  Flow=c(3,6,3,7,5,4,2),
  CumFlow=c(3,9,12,19,5,9,11),
  Med_C=c(10,10,10,10,6,6,6)
)
df

我需要这个来返回这样的表:

id_h  blk_h  blk_w  dist  Flow  CumFlow  Med_C
A     A2     W3     7.0   3     12       10
B     B2     W2     6.5   4     9        6

以下是我为实现这一目标所做的一些尝试:
尝试＃1

library(dplyr)
df.g <- group_by(df, id_h)
df.g2 <- filter(df.g, CumFlow == which.min(CumFlow >= Med_C))

尝试＃2

library(data.table)
setDT(df)[, .SD[which.min(CumCount >= Med_C)], by = id_h]

尝试＃3

library(dplyr)
test <- df %>% group_by(id_h) %>% filter(min(CumFlow) >= Med_C)

我想我误会了如何使用which.min函数。任何意见是极大的赞赏。

最佳答案

两个filter调用可以解决此问题。

使用group_by在每个id_h中工作时，第一个filter返回data.frame，其中CumFlow大于或等于Med_C的所有行。第二个filter在每个id_h中返回CumFlow最低的行。这仅是因为数据已排序。为了使工作更加健壮，您可以考虑在对arrange的调用之后添加对group_by的调用。

library(dplyr)

df <- data.frame(
  id_h    = c("A","A","A","A","B","B","B"),
  blk_h   = c("A1","A1","A2","A2","B1","B2","B2"),
  blk_w   = c("W1","W2","W3","W3","W1","W2","W2"),
  dist    = c(4.3,5.6,7.0,8.7,5.2,6.5,6.8),
  Flow    = c(3,6,3,7,5,4,2),
  CumFlow = c(3,9,12,19,5,9,11),
  Med_C   = c(10,10,10,10,6,6,6)
)
df

df %>%
group_by(id_h) %>%
filter(CumFlow >= Med_C) %>%
filter(CumFlow == min(CumFlow))

关于r - 在x首先超过y的组中过滤，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/38837909/