R-在两个数据帧中连续比较行并返回一个值

我有以下两个数据帧：

df1 <- data.frame(month=c("1","1","1","1","2","2","2","3","3","3","3","3"),
             temp=c("10","15","16","25","13","17","20","5","16","25","30","37"))


df2 <-  data.frame(period=c("1","1","1","1","1","1","1","1","2","2","2","2","2","2","3","3","3","3","3","3","3","3","3","3","3","3"),
              max_temp=c("9","13","16","18","30","37","38","39","10","15","16","25","30","32","8","10","12","14","16","18","19","25","28","30","35","40"),
              group=c("1","1","1","2","2","2","3","3","3","3","4","4","5","5","5","5","5","6","6","6","7","7","7","7","8","8"))

---

我一直在玩

ifelse

函数，需要一些帮助或重新指导。谢谢

如前所述，我发现计算新组的过程很难遵循。据我所知，您试图在

df1

中创建一个名为

new\u group

的变量。对于

df1

的

i行

，

new_组

值是

df2

中第一行的

组

值：

索引为

i

或更高

有一个

期间

值匹配

df1$month[i]

有一个

max_temp

值不小于

df1$temp[i]

我通过对

df1

行索引调用

sapply

来实现这一点：

fxn = function(idx) {
  # Potentially matching indices in df2
  pm = idx:nrow(df2)

  # Matching indices in df2
  m = pm[df2$period[pm] == df1$month[idx] &
         as.numeric(as.character(df1$temp[idx])) <=
         as.numeric(as.character(df2$max_temp[pm]))]

  # Return the group associated with the first matching index
  return(df2$group[m[1]])
}
df1$new_group = sapply(seq(nrow(df1)), fxn)
df1
#    month temp new_group
# 1      1   10         1
# 2      1   15         1
# 3      1   16         1
# 4      1   25         2
# 5      2   13         3
# 6      2   17         4
# 7      2   20         4
# 8      3    5         5
# 9      3   16         6
# 10     3   25         7
# 11     3   30         7
# 12     3   37         8

fxn=函数（idx）{
#df2中的潜在匹配索引
pm=idx:nrow（df2）
#df2中的匹配索引
m=pm[df2$period[pm]==df1$month[idx]&
as.numeric（as.character（df1$temp[idx]））
library（data.table）
dt1您是否有意将数据作为字符串？数据文件实际上是以制表符分隔的文本文件，我使用read.table将其作为数据帧上载到R中。作为R新手，我不知道数据是以字符串形式存在的。数字周围的引号告诉您有字符串。另外，请注意字符串伪装为因子，您将使用我的s
read.table（..stringsAsFactors=TRUE）
（这是默认值）我给出的data.frame代码只是试图重现我在R.Cheers中使用的数据帧。谢谢你的洞察力！谢谢你的有用代码。new_group的值不是计算出来的，只是放在df1$new_group列中的df2$group的值。希望这能让它更清晰。谢谢。是的，我发布的代码就是这样做的。我看起来你对SO还很陌生，没有接受你之前问题的好答案。如果我的解决方案或@RicardoSaporta的解决方案解决了你的问题，记得选中绿色复选框接受它。
fxn = function(idx) {
  # Potentially matching indices in df2
  pm = idx:nrow(df2)

  # Matching indices in df2
  m = pm[df2$period[pm] == df1$month[idx] &
         as.numeric(as.character(df1$temp[idx])) <=
         as.numeric(as.character(df2$max_temp[pm]))]

  # Return the group associated with the first matching index
  return(df2$group[m[1]])
}
df1$new_group = sapply(seq(nrow(df1)), fxn)
df1
#    month temp new_group
# 1      1   10         1
# 2      1   15         1
# 3      1   16         1
# 4      1   25         2
# 5      2   13         3
# 6      2   17         4
# 7      2   20         4
# 8      3    5         5
# 9      3   16         6
# 10     3   25         7
# 11     3   30         7
# 12     3   37         8

library(data.table)
dt1 <- data.table(df1, key="month")
dt2 <- data.table(df2, key="period")

## add a row index
dt1[, rn1 := seq(nrow(dt1))]

dt3 <- 
unique(dt1[dt2, allow.cartesian=TRUE][, new_group := group[min(which(temp <= max_temp))], by="rn1"], by="rn1")

## Keep only the columns you want
dt3[, c("month", "temp", "max_temp", "new_group"), with=FALSE]

    month temp max_temp new_group
 1:     1    1       19         1
 2:     1    3       19         1
 3:     1    4       19         1
 4:     1    7       19         1
 5:     2    2        1         3
 6:     2    5        1         3
 7:     2    6        1         4
 8:     3   10       18         5
 9:     3    4       18         5
10:     3    7       18         5
11:     3    8       18         5
12:     3    9       18         5