Question

我有一个带有以下变量的数据框：

date=c("30/03/2018","30/03/2018","30/03/2018","30/03/2018","30/03/2018","30/03/2018","30/03/2018","30/03/2018")
hour=c(1,1,1,1,2,2,2,2)
location=c(North,South,East,West,North,South,East,West)
North=c(10,30,40,50,0,40,31,11)
South=c(20,10,20,0,0,0,5,0)
East=c(0,10,10,5,0,0,5,0)
West=c(5,30,40,50,0,40,31,11)
df <- data.frame(date, hour, location, North, South, East, West)

location代表观察结果的位置。 North，South，East和West列显示了这些地区的平均降雨量。首先，我需要创建一列Actual来描述location中的降雨量。例如，对于第1行，Actual的值将为10，因为它位于北方。接下来，我需要创建另外两列High1和High2。前者代表该小时内其余3个地区的最高降雨量，后者代表其余3个地区的第二高降雨量。例如，对于第1行，High1和High2的值分别为20和5，因为在该小时中，南方和西方的记录值最高，第二高。

是否存在可以用于此设置的推荐命令？谢谢。

Answer 1

尝试一下（我相信有很多更简单的解决方案）：

        name        a       b
0       anthony    10       5
1       marcus     25       0
2       paul        0       0
3       Aaron     200       7

数据

import pandas as pd
from io import StringIO

s = '''\
name,a,b
anthony,10,5
marcus,75,-50
paul,100,-100
Aaron,200,7
'''
df = pd.read_csv(StringIO(s))

Answer 2

您可以使用以下代码，该代码使用 Rfast 包中的nth函数来获取其余地区的第二高降雨量。

library(Rfast)

NSEW <- grep("North|South|East|West", names(df))
i.col <- sapply(df$location, function(x) grep(x, names(df)))
df$Actual <- as.numeric(df[cbind(1:nrow(df), i.col)])   
df[cbind(1:nrow(df), i.col)] <- 0         
df$High1 <- apply(df[,NSEW], 1, max) # *       
df$High2 <- apply(df[,NSEW], 1, Rfast::nth, k=2, descending=TRUE) # *
df[cbind(1:nrow(df), i.col)] <- df$Actual   
df

        date hour location North South East West Actual High1 High2
1 30/03/2018    1    North    10    20    0    5     10    20     5
2 30/03/2018    1    South    30    10   10   30     10    30    30
3 30/03/2018    1     East    40    20   10   40     10    40    40
4 30/03/2018    1     West    50     0    5   50     50    50     5
5 30/03/2018    2    North     0     0    0    0      0     0     0
6 30/03/2018    2    South    40     0    0   40      0    40    40
7 30/03/2018    2     East    31     5    5   31      5    31    31
8 30/03/2018    2     West    11     0    0   11     11    11     0

*也可以使用rownth，它返回每一行的第n个最小值：

df$High1 <- Rfast::rownth(as.matrix(df[,NSEW]), elems=rep(1, nrow(df)), descending=TRUE)
df$High2 <- Rfast::rownth(as.matrix(df[,NSEW]), elems=rep(2, nrow(df)), descending=TRUE)

Answer 3

如果您不想使用tidyverse，可以选择以下方法：

date <- c("30/03/2018","30/03/2018","30/03/2018","30/03/2018","30/03/2018","30/03/2018","30/03/2018","30/03/2018")
hour <- c(1,1,1,1,2,2,2,2)
loc <- c("North","South","East","West","North","South","East","West")
North <- c(10,30,40,50,0,40,31,11)
South <- c(20,10,20,0,0,0,5,0)
East <- c(0,10,10,5,0,0,5,0)
West <- c(5,30,40,50,0,40,31,11)
location <- list(North, South, East, West, North, South, East, West)



actual <- numeric(length(location))
High1 <- numeric(length(location))
High2 <- numeric(length(location))
for (i in 1:length(location)){
  actual[i] <- (location[[i]][i])
  location[[i]][i] <- 0
  High1[i] <- max(sapply(location , `[` , i))
}
short_location <- location[c(1:4)]
for (i in 1:length(location)){
  n <- length(short_location)
  High2[i] <- sort(sapply(short_location , `[` , i), partial=n-1)[n-1]
}
df <- data.frame(date, hour, loc, North, South, East, West, actual, High1, High2)

将变量名称与R中变量的值匹配

3 个答案: