Question

我有一个大型数据集，我使用dplyr() summarize来生成一些方法。偶尔，我想对该输出执行算术运算。例如，我想从下面的输出中得到平均值，比如＆＃34; m.biomass＆＃34;。

我已尝试过此mean(data.sum[,7])和此mean(as.list(data.sum[,7]))。有没有快速简便的方法来实现这一目标？

data.sum <-structure(list(scenario = c("future", "future", "future", "future"
), state = c("fl", "ga", "ok", "va"), m.soc = c(4090.31654013689, 
3654.45350562628, 2564.33199749487, 4193.83388887064), m.npp = c(1032.244475, 
821.319385, 753.401315, 636.885535), sd.soc = c(56.0344229400332, 
97.8553643582118, 68.2248389927858, 79.0739969429246), sd.npp = c(34.9421782033153, 
27.6443555578531, 26.0728757486901, 24.0375040705595), m.biomass = c(5322.76631158111, 
3936.79457763176, 3591.0902359206, 2888.25308402464), sd.m.biomass = c(3026.59250918009, 
2799.40317348016, 2515.10516340438, 2273.45510178843), max.biomass = c(9592.9303, 
8105.109, 7272.4896, 6439.2259), time = c("1980-1999", "1980-1999", 
"1980-1999", "1980-1999")), .Names = c("scenario", "state", "m.soc", 
"m.npp", "sd.soc", "sd.npp", "m.biomass", "sd.m.biomass", "max.biomass", 
"time"), class = c("grouped_df", "tbl_df", "tbl", "data.frame"
), row.names = c(NA, -4), vars = list(quote(scenario)), labels = structure(list(
    scenario = "future"), class = "data.frame", row.names = c(NA, 
-1), vars = list(quote(scenario)), drop = TRUE, .Names = "scenario"), indices = list(0:3))

Answer 1

我们可以使用[[将列提取为vector;由于mean仅适用于vector或matrix - 而非data.frame。如果OP希望在单个列上执行此操作，请使用：

mean(data.sum[[7]]) 
#[1] 3934.726

如果只有data.frame课程，则data.sum[,7]会将其解除为vector，但tbl_df会阻止其将其展开为vector }

对于多列，dplyr也有专门的功能

data.sum %>%
       summarise_each(funs(mean), 3:7)

来自R中dplyr的汇总数据帧的算法

1 个答案: