我使用：
加载了一个csv文件data = read.csv(file="/home/stefanos/R/data_frames_new/temp2.csv", header=TRUE, sep=",")
temp2.csv文件的前4行是：
nodeId,battery,date,idHistoric,temperature,longitude,latitude
3,78.00,2013-04-01 00:04:03,30163676,13.74,-3.80176,43.46192
3,78.00,2013-04-01 00:09:01,30164278,13.67,-3.80176,43.46192
3,78.00,2013-04-01 00:13:59,30164875,13.67,-3.80176,43.46192
我想通过nodeId对其进行分组，并找到每15分钟的温度平均值。所以我输入：
df <- xts(x = data[, c("nodeId", "battery", "idHistoric", "temperature", "longitude", "latitude")], order.by = as.POSIXct(data[, "date"], tz = "GMT", format = "%Y-%m-%d %H:%M:%S"))
然后：
df2 <- by(df,df$nodeId,function(x){
ends <- endpoints(x, on = "minutes", k = 15)
xx <- period.apply(x, ends, mean)
})
我的问题是我无法将df2写入csv文件。我还没有能够这样做。 当我在屏幕上打印df2时，我看到以下结构：
/*********************************************/
INDICES: 3
nodeId battery idHistoric temperature longitude latitude
2013-04-01 00:13:59 3 78.00000 30164276 13.69333 -3.80176 43.46192
2013-04-01 00:28:54 3 79.00000 30166075 13.78000 -3.80176 43.46192
[...]
------------------------------------------------------------
INDICES: 4
nodeId battery idHistoric temperature longitude latitude
2013-04-01 00:13:07 4 87.00000 30164172 14.42667 -3.80098 43.46199
2013-04-01 00:28:01 4 87.33333 30165964 14.49000 -3.80098 43.46199
------------------------------------------------------------
INDICES: 5
nodeId battery idHistoric temperature longitude latitude
2013-04-01 00:13:31 5 83.00000 30164224 13.84667 -3.80058 43.46203
2013-04-01 00:28:26 5 83.66667 30166018 14.06000 -3.80058 43.46203
------------------------------------------------------------
INDICES: 6
nodeId battery idHistoric temperature longitude latitude
2013-04-01 00:12:52 6 78.00000 30164128 13.99667 -3.79979 43.46212
2013-04-01 00:28:52 6 79.00000 30165983 13.97333 -3.79979 43.46212
/*********************************************/
那么如何将其保存为CSV？
答案 0 :(得分：3)
您可以这样做（如评论中@Roland所述）
write.table(do.call(rbind,df2),file='test.csv')
这是一个包含数据的完整示例。您可以使用read.zoo在一个衬管命令中创建xts对象：
library(zoo)
## you replace text= here by file=temp2.csv
dat <- read.zoo(text='nodeId,battery,date,idHistoric,temperature,longitude,latitude
3,78.00,2013-04-01 00:13:59,30163676,13.74,-3.80176,43.46192
3,78.00,2013-04-01 00:28:54,30163676,13.74,-3.80176,43.46192
4,78.00,2013-04-01 00:13:07,30164278,13.67,-3.80176,43.46192
4,78.00,2013-04-01 00:28:01,30163676,13.74,-3.80176,43.46192
5,78.00,2013-04-01 00:13:31,30163676,13.74,-3.80176,43.46192
5,78.00,2013-04-01 00:28:26,30164875,13.67,-3.80176,43.46192
6,78.00,2013-04-01 00:12:52,30164875,13.67,-3.80176,43.46192
6,78.00,2013-04-01 00:28:52,30164875,13.67,-3.80176,43.46192',header=TRUE,
tz='',sep=',',index=3)
然后按组创建并保存列表
library(xts)
df2 <- by(dat,dat$nodeId,function(x){
ends <- endpoints(x, on = "minutes", k = 1)
xx <- period.apply(x, ends, mean)
})
write.table(do.call(rbind,df2),file='test.csv')
再次阅读，你只需
read.table('test.csv')
nodeId battery idHistoric temperature longitude latitude
3.2013-04-01 00:13:59 3 78 30163676 13.74 -3.80176 43.46192
3.2013-04-01 00:28:54 3 78 30163676 13.74 -3.80176 43.46192
4.2013-04-01 00:13:07 4 78 30164278 13.67 -3.80176 43.46192
4.2013-04-01 00:28:01 4 78 30163676 13.74 -3.80176 43.46192
5.2013-04-01 00:13:31 5 78 30163676 13.74 -3.80176 43.46192
5.2013-04-01 00:28:26 5 78 30164875 13.67 -3.80176 43.46192
6.2013-04-01 00:12:52 6 78 30164875 13.67 -3.80176 43.46192
6.2013-04-01 00:28:52 6 78 30164875 13.67 -3.80176 43.46192
编辑以保存/再次将其作为动物园对象进行阅读，我略微改变了绑定列表的rownames：
dd <- do.call(rbind,df2)
rownames(dd) <- gsub('*.[.]','',rownames(dd))
write.table(dd,file='test.csv')
现在我可以再读一遍：
read.zoo('test.csv',index=0,tz='')
nodeId battery idHistoric temperature longitude latitude
2013-04-01 00:12:52 6 78 30164875 13.67 -3.80176 43.46192
2013-04-01 00:13:07 4 78 30164278 13.67 -3.80176 43.46192
2013-04-01 00:13:31 5 78 30163676 13.74 -3.80176 43.46192
2013-04-01 00:13:59 3 78 30163676 13.74 -3.80176 43.46192
2013-04-01 00:28:01 4 78 30163676 13.74 -3.80176 43.46192
2013-04-01 00:28:26 5 78 30164875 13.67 -3.80176 43.46192
2013-04-01 00:28:52 6 78 30164875 13.67 -3.80176 43.46192
2013-04-01 00:28:54 3 78 30163676 13.74 -3.80176 43.46192
<强> EDIT2
感谢@Gsee出色的回答，您可以执行以下操作：
do.call(rbind, unname(df2))
这将保持行名称正确，因此不需要像我之前的编辑那样使用正则表达式。
答案 1 :(得分：2)
如果您查看str(df2)，您会看到它是一个名为list的人。通常，如果您有一个列表并且想要将其转换为单个对象，则可以使用do.call(rbind, df2)之类的内容。这与rbind(df2[[1]], df2[[2]], df2[[3]], df2[[4]])相同，但可以使用任意长度的列表。
在这种情况下，您的list有names
> names(df2)
[1] "3" "4" "5" "6"
所以，如果你只是do.call(rbind, df2)，那么rownames将不是您想要的 - 它们将被添加到列表的names之前。
> rownames(do.call(rbind, df2))
[1] "3.2013-04-01 00:13:59" "3.2013-04-01 00:28:54" "4.2013-04-01 00:13:07"
[4] "4.2013-04-01 00:28:01" "5.2013-04-01 00:13:31" "5.2013-04-01 00:28:26"
[7] "6.2013-04-01 00:12:52" "6.2013-04-01 00:28:52"
解决方案是unname列表
do.call(rbind, unname(df2))
由于您正在使用xts，因此您可能希望将其强制转换为xts对象：
> as.xts(do.call(rbind, unname(df2)))
nodeId battery idHistoric temperature longitude latitude
2013-04-01 00:12:52 6 78 30164875 13.67 -3.80176 43.46192
2013-04-01 00:13:07 4 78 30164278 13.67 -3.80176 43.46192
2013-04-01 00:13:31 5 78 30163676 13.74 -3.80176 43.46192
2013-04-01 00:13:59 3 78 30163676 13.74 -3.80176 43.46192
2013-04-01 00:28:01 4 78 30163676 13.74 -3.80176 43.46192
2013-04-01 00:28:26 5 78 30164875 13.67 -3.80176 43.46192
2013-04-01 00:28:52 6 78 30164875 13.67 -3.80176 43.46192
2013-04-01 00:28:54 3 78 30163676 13.74 -3.80176 43.46192
最后，我发现使用write.zoo编写xts或zoo个对象的csv文件很方便：
write.zoo(as.xts(do.call(rbind, unname(df2))), file="test.csv", sep=",")
答案 2 :(得分：1)
解决
xts返回一个向量。我不得不遍历每个位置（使用while）并将每个元素写入csv文件。
df2 <- by(df,df$nodeId,function(x){
ends <- endpoints(x, on = "minutes", k = 15)
xx <- period.apply(x, ends, mean)
})
i <- 1
total <- length(df2)
while( i <= total ){
write.csv(df2[i],paste("lights_2013-04-0102/out_",i,".csv",sep = ""))
i <- i + 1
}