计算行的方式 [英] Calculate means of rows

问题描述

我有一个名为 ants 的数据框，详细说明每个站点的多个条目，如下所示：

I have a dataframe called ants detailing multiple entries per site, looks like this:

  Site     Date     Time  Temp SpCond Salinity Depth Turbidity Chlorophyll
1   71 6/8/2010 14:50:35 14.32  49.88    32.66 0.397       0.0         1.3
2   71 6/8/2010 14:51:00 14.31  49.94    32.70 1.073       0.0         2.0
3   71 6/8/2010 14:51:16 14.32  49.95    32.71 1.034      -0.1         1.6
4   71 6/8/2010 14:51:29 14.31  49.96    32.71 1.030      -0.2         1.6
5   70 6/8/2010 14:53:55 14.30  50.04    32.77 1.002      -0.2         1.2
6   70 6/8/2010 14:54:09 14.30  50.03    32.77 0.993      -0.5         1.2

网站具有不同数量的条目，通常为3，但有时较少或更多。如果日期和站点号码匹配，我想编写一个新的数据框，每个站点有一个条目，详细说明每个参数的平均值/平均读数。我想要从计算和后续数据框中省略空或na单元格。

Sites have different numbers of entries, usually 3 but sometimes less or more. Where both date and site number match I would like to write a new dataframe with one entry per site detailing the average/mean readings for each parameter. I would like empty or "na" cells to be omitted from the calculation and subsequent dataframe.

我不知道这是一个应用功能还是一个版本的行也许？非常困难，任何帮助非常感谢！

I'm not sure if this is an apply function or a version of rowMeans maybe? Very stuck, any help much appreciated!

推荐答案

这是一个完整的新答案，全面的日志也涵盖了您的新规范： / p>

Here is a complete new answer with a full log also covering your new specification:

R> Lines <- "  Site     Date     Time  Temp SpCond Salinity Depth Turbidity Chlorophyll
+ 71 6/8/2010 14:50:35 14.32  49.88    32.66 0.397       0.0         1.3
+ 71 6/8/2010 14:51:00 14.31  49.94    32.70 1.073       0.0         2.0
+ 71 6/8/2010 14:51:16 14.32  49.95    32.71 1.034      -0.1         1.6
+ 71 6/8/2010 14:51:29 14.31  49.96    32.71 1.030      -0.2         1.6
+ 70 6/8/2010 14:53:55 14.30  50.04    32.77 1.002      -0.2         1.2
+ 70 6/8/2010 14:54:09 14.30  50.03    32.77 0.993      -0.5         1.2
+ "
R> con <- textConnection(Lines)
R> df <- read.table(con, sep="", header=TRUE, stringsAsFactors=FALSE)
R> close(con)
R> df$pt <- as.POSIXct(strptime(paste(df$Date, df$Time), "%m/%d/%Y %H:%M:%S"))
R> library(plyr)
R> newdf <- ddply(df, .(Site,Date), function(x) mean(x[,-(1:3)], na.rm=TRUE))
R> newdf$pt <- as.POSIXct(newdf$pt, origin="1970-01-01")
R> newdf
  Site     Date  Temp SpCond Salinity  Depth Turbidity Chlorophyll                  pt
1   70 6/8/2010 14.30  50.03    32.77 0.9975    -0.350       1.200 2010-06-08 20:54:02
2   71 6/8/2010 14.32  49.93    32.70 0.8835    -0.075       1.625 2010-06-08 20:51:05
R> 

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