R连续连续天有条件

Question

我有下一个问题。如果我有下一个数据框

data<- data.frame(
        Time= c("2018-01-01", "2018-01-02", "2018-01-03", "2018-01-04", "2018-01-05"),
        TEN= c(10,20,11,12,16)
    )

，我想使用将TEN <15的连续天数作为新列。

我尝试

waves_min <- function(df, prop, min_value, min_days) {
    sum(with(rle(df$Temp > min_temp), values & lengths >= min_days))
}

但它返回该计数的总数，而不是每一行的值。

有什么想法吗？

谢谢

Answer 1

@Entity @Table(name = "some_entity") public final class SomeEntity { @Column(nullable = false, unique = true, length = 20) private String externalId; @Column(nullable = false, length = 50) private String name; @ManyToOne(fetch = FetchType.EAGER) @JoinColumn(name = "parentId", nullable = true) private SomeEntity parent; @OneToMany(mappedBy = "parent") private List<SomeEntity> children; //or Set<> //returns direct children public List<SomeEntity> getChildren(){ return children; } //returns all children to any level public List<SomeEntity> getAllChildren(){ getAllChildren(this); } //recursive function to walk the tree private List<SomeEntity> getAllChildren(SomeEntity parent){ List<SomeEntity> allChidren = new ArrayList<>(); for(SomeEntity child : children){ allChildren.add(child); allChildren.addAll(getAllChildren(child); } return allChildren; } } + dplyr：

data.table::rleid

输出：

library(dplyr)
library(data.table)

data %>%
  group_by(ID = data.table::rleid(TEN < 15)) %>%
  mutate(Consec_Days = if_else(TEN < 15, row_number(), 0L))

# A tibble: 7 x 4 # Groups: ID [5] Time TEN ID Consec_Days <fct> <dbl> <int> <int> 1 2018-01-01 10 1 1 2 2018-01-02 20 2 0 3 2018-01-03 11 3 1 4 2018-01-04 12 3 2 5 2018-01-05 16 4 0 6 2018-01-06 17 4 0 7 2018-01-07 14 5 1 ：

data.table

输出：

setDT(data)
data[, Consec_Days := ifelse(TEN < 15, 1:.N, 0L), by = rleid(TEN < 15)]

基本R + Time TEN Consec_Days 1: 2018-01-01 10 1 2: 2018-01-02 20 0 3: 2018-01-03 11 1 4: 2018-01-04 12 2 5: 2018-01-05 16 0 6: 2018-01-06 17 0 7: 2018-01-07 14 1 ：

data.table::rleid

输出：

data$Consec_Days <- with(data, ave(TEN, data.table::rleid(TEN < 15), 
                                   FUN = function(x) ifelse(x < 15, seq_along(x), 0L)))

数据：

        Time TEN Consec_Days
1 2018-01-01  10           1
2 2018-01-02  20           0
3 2018-01-03  11           1
4 2018-01-04  12           2
5 2018-01-05  16           0
6 2018-01-06  17           0
7 2018-01-07  14           1

我在OP的示例数据中添加了更多行，以说明这些解决方案适用于所有情况。