【问题标题】:Calculate mean for each row across a list of dataframes in R计算 R 中数据框列表中每一行的平均值
【发布时间】:2020-04-02 19:39:51
【问题描述】:

我想计算数据框列表中“值”列中每一行的平均值。

我想要的输出是一个数据框,其中包含整个列表中每一行的“样本”名称及其相关的平均“值”。

以下是一些示例数据:

list <- list()

dataframe1 <- data.frame(sample = c("OP2645ii_c","OP5048___e","OP5048___f","OP5046___d","OP2645ii_e","OP2645ii_a","OP5054DNAa","OP5048___c","OP2645ii_d","OP5048___b","OP5047___a","OP5048___h","OP5053DNAb","OP3088i__a","OP5048___g","OP5053DNAa","OP5049___a","OP2645ii_b","OP5046___c","OP5044___c","OP2413iiia","OP5054DNAc","OP5046___e","OP5054DNAb","OP5044___a","OP5046___a","OP5046___b","OP2413iiib","OP5051DNAa","OP5048___d","OP5044___b","OP5049___b","OP5051DNAc","OP5051DNAb","OP5053DNAc","OP5047___b","OP5043___b","OP5043___a","OP5052DNAa"),
                         gr = c("1","1","2","5","4","5","5","3","2","2","2","4","3","1","1","3","2","1","2","5","5","5","2","2","2","1","2","1","1","1","2","1","1","2","2","5","3","3","5"),
                         value = c("14.32500","14.32500","14.32500","14.32500","14.32500","14.32500","14.32500","14.32500","14.32500","14.32500","14.32500","14.32500","14.32500","14.32500","14.32500","14.32500","14.32500","14.32500","14.32500","14.32500","14.32500","14.32500","14.32500","14.32500","14.32500","14.32500","14.32500","14.32500","14.32500","14.32500","14.32500","14.32500","14.32500","14.32500","14.32500","14.32500","14.32500","14.32500","14.32500"))

list[[1]] <- dataframe1

dataframe2 <- data.frame(sample = c("OP2645ii_c","OP5048___e","OP5048___f","OP5046___d","OP2645ii_e","OP2645ii_a","OP5054DNAa","OP5048___c","OP2645ii_d","OP5048___b","OP5047___a","OP5048___h","OP5053DNAb","OP3088i__a","OP5048___g","OP5053DNAa","OP5049___a","OP2645ii_b","OP5046___c","OP5044___c","OP2413iiia","OP5054DNAc","OP5046___e","OP5054DNAb","OP5044___a","OP5046___a","OP5046___b","OP2413iiib","OP5051DNAa","OP5048___d","OP5044___b","OP5049___b","OP5051DNAc","OP5051DNAb","OP5053DNAc","OP5047___b","OP5043___b","OP5043___a","OP5052DNAa"),
           gr = c("5","4","3","5","4","5","5","3","2","2","2","2","3","1","1","3","2","1","2","5","5","5","2","2","2","1","2","1","1","1","2","1","1","2","4","4","4","4","4"),
           value = c("12.59000","12.59000","12.59000","12.59000","12.59000","12.59000","12.59000","12.59000","12.59000","12.59000","12.59000","12.59000","12.59000","12.59000","12.59000","12.59000","12.59000","12.59000","12.59000","12.59000","12.59000","12.59000","12.59000","12.59000","12.59000","12.59000","12.59000","12.59000","12.59000","12.59000","12.59000","12.59000","12.59000","12.59000","12.59000","12.59000","12.59000","12.59000","12.59000"))
list[[2]] <- dataframe2

dataframe3 <- data.frame(sample = c("OP2645ii_c","OP5048___e","OP5048___f","OP5046___d","OP2645ii_e","OP2645ii_a","OP5054DNAa","OP5048___c","OP2645ii_d","OP5048___b","OP5047___a","OP5048___h","OP5053DNAb","OP3088i__a","OP5048___g","OP5053DNAa","OP5049___a","OP2645ii_b","OP5046___c","OP5044___c","OP2413iiia","OP5054DNAc","OP5046___e","OP5054DNAb","OP5044___a","OP5046___a","OP5046___b","OP2413iiib","OP5051DNAa","OP5048___d","OP5044___b","OP5049___b","OP5051DNAc","OP5051DNAb","OP5053DNAc","OP5047___b","OP5043___b","OP5043___a","OP5052DNAa"),
                         gr = c("5","3","3","5","5","5","5","3","5","3","3","3","3","3","3","3","2","1","2","1","1","1","2","2","2","1","2","1","1","1","2","1","1","4","4","4","4","4","4"),
                         value = c("20.06915","20.06915","20.06915","20.06915","20.06915","20.06915","20.06915","20.06915","20.06915","20.06915","20.06915","20.06915","20.06915","20.06915","20.06915","20.06915","20.06915","20.06915","20.06915","20.06915","20.06915","20.06915","20.06915","20.06915","20.06915","20.06915","20.06915","20.06915","20.06915","20.06915","20.06915","20.06915","20.06915","20.06915","20.06915","20.06915","20.06915","20.06915","20.06915"))

list[[3]] <- dataframe3

dataframe4 <- data.frame(sample = c("OP2645ii_c","OP5048___e","OP5048___f","OP5046___d","OP2645ii_e","OP2645ii_a","OP5054DNAa","OP5048___c","OP2645ii_d","OP5048___b","OP5047___a","OP5048___h","OP5053DNAb","OP3088i__a","OP5048___g","OP5053DNAa","OP5049___a","OP2645ii_b","OP5046___c","OP5044___c","OP2413iiia","OP5054DNAc","OP5046___e","OP5054DNAb","OP5044___a","OP5046___a","OP5046___b","OP2413iiib","OP5051DNAa","OP5048___d","OP5044___b","OP5049___b","OP5051DNAc","OP5051DNAb","OP5053DNAc","OP5047___b","OP5043___b","OP5043___a","OP5052DNAa"),
                         gr = c("2","2","2","3","4","5","5","3","2","2","2","4","5","1","1","3","2","1","2","5","5","5","2","2","2","1","2","1","1","1","2","1","1","2","2","5","3","3","5"),
                         value = c("18.32500","18.32500","18.32500","18.32500","18.32500","18.32500","18.32500","18.32500","18.32500","18.32500","18.32500","18.32500","18.32500","18.32500","18.32500","18.32500","18.32500","18.32500","18.32500","18.32500","18.32500","18.32500","18.32500","18.32500","18.32500","18.32500","18.32500","18.32500","18.32500","18.32500","18.32500","18.32500","18.32500","18.32500","18.32500","18.32500","18.32500","18.32500","18.32500"))

list[[4]] <- dataframe4

非常感谢。 干杯。 迪翁

【问题讨论】:

    标签: r list dataframe mean


    【解决方案1】:

    使用base 函数,您可以将所有value 列提取到一个矩阵中并使用行表示:

    rowMeans(sapply(list, "[[", "value"))
    

    对于您的样本数据,您还需要转换为数字(如下所示),但我希望您的真实数据包含数字而不是因子。

    rowMeans(sapply(lapply(list, "[[", "value"), function(x) as.numeric(as.character(x))))
    

    这只是给出了值(并假设行的顺序正确)。您可以使用cbind 添加示例名称,例如cbind(list[[1]][["sample"]], rowMeans(...))

    【讨论】:

      【解决方案2】:

      我们可以将list 元素绑定到单个data.frame,创建一组row_number() 并使用它来获取'value'的mean

      library(dplyr)
      bind_rows(list, .id = 'id') %>% 
                mutate(value = as.numeric(value))  %>%
                group_by(id) %>% 
                group_by(grp = row_number(), sample) %>%
                summarise(value = mean(value, na.rm = TRUE))
      

      【讨论】:

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