0

我需要回答

no fac date count

1 fac1 2015-01 3
2 fac1 2016-01 5
3 fac1 2017-01 7
4 fac2 2015-01 9
5 fac2 2017-01 11
6 fac3 2016-01 13
7 fac4 2015-01 15
8 fac5 2017-01 17

date     fac1 fac2 fac3 fac4 fac5
2015-01     3    9    0   15    0
2016-01     5    0   13    0    0
2017-01     7   11    0    0   17

像这样。

我正在尝试使用左连接,但会发生此错误。

# first Data
Data0 <-mid[1:31,]

# Separate Data by factor
for (i in 1:72) {
  assign(paste0("Data", i), subset(mid, midnames %in% mid_names[i])) 
}

# Make first dataset
df3 <- Data0
df3$date[is.na(df3$date)] <- Sys.time()
df3[is.na(df3)]<-0

#leftjoin Datas by factor
for (i in 1:72) {
  df3 <- sqldf(paste("SELECT *
                FROM df3
                   LEFT OUTER JOIN Data",i," USING(year,month,date)",sep=""))
  df3[is.na(df3)]<-0

}

错误:无法将 NA 传递给 dbQuoteIdentifier()

另外:警告消息:在 field_types[] <- field_types[names(data)] 中:要替换的项目数不是替换长度的倍数

4

1 回答 1

0

您的查询已关闭;你需要一个标准的数据透视查询:

SELECT
    date,
    MAX(CASE WHEN fac = 'fac1' THEN count END) AS fac1,
    MAX(CASE WHEN fac = 'fac2' THEN count END) AS fac2,
    MAX(CASE WHEN fac = 'fac3' THEN count END) AS fac3,
    MAX(CASE WHEN fac = 'fac4' THEN count END) AS fac4,
    MAX(CASE WHEN fac = 'fac5' THEN count END) AS fac5
FROM yourTable
GROUP BY
    date

我假设您有一个数据框,其中包含您的第一条样本信息中的数据。

于 2017-11-06T07:08:08.743 回答