将代码应用于24个相似数据集的最佳方式？
将代码应用于24个相似数据集的最佳方式？
提问于 2019-08-03 02:37:47
我有24个数据集，每个数据集都有一个因子和一个响应。我已经编写了将93个条目分成3个类别的代码，但我不确定为我的所有24个数据集运行此代码的最有效方法是什么。任何想法都将不胜感激。
这是我正在处理的数据。
dput(head(data))
structure(list(run.size.percentage = structure(c(2L, 13L, 24L,
35L, 46L, 57L), .Label = c(",2000,", "1,0.375,0.013", "10,0.868,0.11",
"11,0.953,0.12", "12,1.047,0.12", "13,1.149,0.13", "14,1.261,0.14",
"15,1.385,0.14", "16,1.520,0.15", "17,1.668,0.15", "18,1.832,0.16",
"19,2.011,0.17", "2,0.412,0.023", "20,2.207,0.17", "21,2.423,0.18",
"22,2.660,0.19", "23,2.920,0.20", "24,3.205,0.21", "25,3.519,0.22",
"26,3.863,0.24", "27,4.240,0.25", "28,4.655,0.26", "29,5.110,0.28",
"3,0.452,0.034", "30,5.610,0.30", "31,6.158,0.31", "32,6.760,0.33",
"33,7.421,0.35", "34,8.147,0.37", "35,8.943,0.39", "36,9.817,0.42",
"37,10.78,0.45", "38,11.83,0.47", "39,12.99,0.50", "4,0.496,0.049",
"40,14.26,0.53", "41,15.65,0.56", "42,17.18,0.58", "43,18.86,0.59",
"44,20.70,0.59", "45,22.73,0.58", "46,24.95,0.55", "47,27.39,0.52",
"48,30.07,0.49", "49,33.01,0.46", "5,0.545,0.061", "50,36.24,0.45",
"51,39.78,0.45", "52,43.67,0.45", "53,47.94,0.44", "54,52.62,0.42",
"55,57.77,0.38", "56,63.41,0.35", "57,69.61,0.32", "58,76.42,0.31",
"59,83.89,0.33", "6,0.598,0.072", "60,92.09,0.36", "61,101.1,0.42",
"62,111.0,0.49", "63,121.8,0.59", "64,133.7,0.74", "65,146.8,0.94",
"66,161.2,1.19", "67,176.9,1.49", "68,194.2,1.82", "69,213.2,2.18",
"7,0.656,0.083", "70,234.1,2.55", "71,256.9,2.94", "72,282.1,3.34",
"73,309.6,3.78", "74,339.9,4.25", "75,373.1,4.73", "76,409.6,5.20",
"77,449.7,5.60", "78,493.6,5.87", "79,541.9,5.93", "8,0.721,0.093",
"80,594.9,5.77", "81,653.0,5.37", "82,716.8,4.77", "83,786.9,4.03",
"84,863.9,3.21", "85,948.3,2.36", "86,1041,1.55", "87,1143,0.81",
"88,1255,0.30", "89,1377,0.056", "9,0.791,0.10", "90,1512,0.0044",
"91,1660,0", "92,1822,0"), class = "factor")), row.names = c(NA,
6L), class = "data.frame")下面是为每个数据集工作的代码。
data2 <- tidyr::separate(names(data), unlist(strsplit(names(data), "\\.")), ",", data=data)
group1 <- data2 %>% filter(size <= 2)
group2 <- data2 %>% filter(size > 2 & size <= 50)
group3 <- data2 %>% filter(size > 50 & size <= 2000)
sum(as.numeric(group1$percentage), na.rm=TRUE)
sum(as.numeric(group2$percentage), na.rm=TRUE)
sum(as.numeric(group3$percentage), na.rm=TRUE)回答 1
Stack Overflow用户
发布于 2019-08-03 03:00:29
将你的数据帧放在一个列表中，并使用lapply。已使用cut创建所需的大小组。我还添加了convert = TRUE参数到separate，以将数字转换为数字-
df_list <- list(df, df) # creating a dummy list with same df
lapply(df_list, function(x) {
separate(names(df), unlist(strsplit(names(df), "\\.")), ",",
data = df, convert = TRUE) %>%
group_by(group = cut(size, breaks = c(0,2,50,2000,Inf))) %>%
summarise(percentage = sum(percentage))
})
# every list element is your desired output df
[[1]]
# A tibble: 1 x 2
group percentage
<fct> <dbl>
1 (0,2] 0.252
[[2]]
# A tibble: 1 x 2
group percentage
<fct> <dbl>
1 (0,2] 0.252页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/57331880
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