我有两个csv文件:
文件1:
SN CY Year Month Day Hour Lat Lon
196101 1 1961 1 14 12 8.3 134.7
196101 1 1961 1 14 18 8.8 133.4
196101 1 1961 1 15 0 9.1 132.5
196101 1 1961 1 15 6 9.3 132.2
196101 1 1961 1 15 12 9.5 132
196101 1 1961 1 15 18 9.9 131.8
file2的:
Year Month Day RR Hour Lat Lon
1961 1 14 0 0 14.0917 121.055
1961 1 14 0 6 14.0917 121.055
1961 1 14 0 12 14.0917 121.055
1961 1 14 0 18 14.0917 121.055
1961 1 15 0 0 14.0917 121.055
1961 1 15 0 6 14.0917 121.055
我想在file2中添加另一列并放入" TRUE"如果file2中的行存在于file1中,只要它们具有相同的年,月,日和小时,否则" FALSE"。然后另存为csv文件。
所需的输出:
Year Month Day RR Hour Lat Lon com
1961 1 14 0 0 14.0917 121.055 FALSE
1961 1 14 0 6 14.0917 121.055 FALSE
1961 1 14 0 12 14.0917 121.055 TRUE
1961 1 14 0 18 14.0917 121.055 TRUE
1961 1 15 0 0 14.0917 121.055 TRUE
1961 1 15 0 6 14.0917 121.055 TRUE
这是我的剧本:
jtwc <- read.csv("file1.csv",header=T,sep=",")
stn <- read.csv("file2.csv",header=T,sep=",")
if ((jtwc$Year == "stn$YY") & (jtwc$Month == "stn$MM") & (jtwc$Day == "stn$DD") &(jtwc$Hour == "stn$HH")){
stn$com <- "TRUE"
} else {
stn$com <- "FALSE"
}
write.csv(stn,file="test.csv",row.names=T)
这会出错:
In if ((jtwc$Year == "stn$YY") & (jtwc$Month == "stn$MM") & (jtwc$Day == :the condition has length > 1 and only the first element will be used
答案 0 :(得分:3)
您也可以使用dplyr / tidyverse:
library(tidyverse)
d2 %>%
left_join(select(d1, Year, Month, Day, Hour, Com=Lon)) %>%
mutate(Com=ifelse(is.na(Com), FALSE, TRUE))
Joining, by = c("Year", "Month", "Day", "Hour")
Year Month Day RR Hour Lat Lon Com
1 1961 1 14 0 0 14.0917 121.055 FALSE
2 1961 1 14 0 6 14.0917 121.055 FALSE
3 1961 1 14 0 12 14.0917 121.055 TRUE
4 1961 1 14 0 18 14.0917 121.055 TRUE
5 1961 1 15 0 0 14.0917 121.055 TRUE
6 1961 1 15 0 6 14.0917 121.055 TRUE
答案 1 :(得分:1)
使用data.table
快速而肮脏的解决方案:
fread
读取文件。file1
中提取所需的列(因为您只对file2
感兴趣)merge
file1
添加FALSE
代码:
library(data.table)
result <- merge(fread("file2.csv"),
fread("file1.csv")[, .(Year, Month, Day, Hour, com = TRUE)],
all.x = TRUE)[is.na(com), com := FALSE]
result
Year Month Day Hour RR Lat Lon com
1: 1961 1 14 0 0 14.0917 121.055 FALSE
2: 1961 1 14 6 0 14.0917 121.055 FALSE
3: 1961 1 14 12 0 14.0917 121.055 TRUE
4: 1961 1 14 18 0 14.0917 121.055 TRUE
5: 1961 1 15 0 0 14.0917 121.055 TRUE
6: 1961 1 15 6 0 14.0917 121.055 TRUE