Question

我的桌子看起来像这样：

DT <-data.table(ID=c(1:4),AREA=c("a","b","c","d"),PARTNER=c("f","b","g","d"),OBS_VALUE=c(10,5,13,0))

因此，我想获得AREA和PARTNER相等且OBS_VALUE不等于0或NA的记录。

如果两列相同，则全局相同的函数测试。

setkeyv(DT,c("AREA","PARTNER"))
identical(DT['AREA'],DT['PARTNER'])

结果显然是假的。

我不知道如何抵达目标。谢谢你的帮助。

收到答复

DT[REF_AREA==COUNTERPART_AREA & !is.na(OBS_VALUE) & OBS_VALUE!=0]

给出错误消息：

Error in Ops.factor(REF_AREA, COUNTERPART_AREA) : 
  level sets of factors are different

确实是我的数据。表格更复杂：

dput(head(diagonal))
structure(list(TIME_PERIOD = c(2010L, 2010L, 2010L, 2010L, 2010L, 
2010L), REF_AREA = structure(c(1L, 1L, 1L, 1L, 1L, 1L), .Label = c("AT", 
"BE", "BG", "CY", "CZ", "DE", "DK", "EE", "ES", "FI", "FR", "GB", 
"GR", "HR", "HU", "IE", "IT", "LT", "LU", "LV", "MT", "NL", "PL", 
"PT", "RO", "SE", "SI", "SK"), class = "factor"), COUNTERPART_AREA =     structure(c(20L, 20L, 20L, 20L, 20L, 20L), .Label = c("4A", "4F", "4S", "9A", 
"A1", "A2", "A5", ... "AT", ..., "W1"), class = "factor"), 
, .Names = c("TIME_PERIOD", "REF_AREA", "COUNTERPART_AREA", "UNIT_MEASURE", "INT_ACC_ITEM", "ACCOUNTING_ENTRY", "OBS_VALUE", "OBS_COMMENT", "DECIMALS", "UNIT_MULT", "i.CONF_STATUS", "i.OBS_STATUS"), sorted = c("REF_AREA", "COUNTERPART_AREA"), class = c("data.table", "data.frame"), row.names = c(NA, -6L), .internal.selfref = <pointer: 0x07d424a0>)

有什么想法吗？

Answer 1

DT[AREA==PARTNER & !is.na(OBS_VALUE) & OBS_VALUE!=0]

Answer 2

使用dplyr：

library(dplyr)

DT <-data.frame(ID=c(1:4),AREA=c("a","b","c","d"),PARTNER=c("f","b","g","d"),OBS_VALUE=c(10,5,13,0), stringsAsFactors = FALSE)

DT %>% 
  filter(AREA == PARTNER & OBS_VALUE != 0 & !is.na(OBS_VALUE))

保留R

2 个答案: