我想将,替换为.作为数据帧中的小数点。我可以使用df$X2005 <- as.numeric(gsub(',', '.', df$X2005))为每个变量执行此操作。是否有更有效的方法可以同时为整个数据框执行此操作？
一些示例数据：
df <- structure(list(country = structure(1:6, .Label = c("Australia", "Austria", "Belgium", "Canada", "Chile", "Czech Republic", "Denmark", "Estonia", "Finland", "France", "Germany", "Greece", "Hungary", "Iceland", "Ireland", "Israel", "Italy", "Japan", "Korea", "Luxembourg", "Mexico", "Netherlands", "New Zealand", "Norway", "Poland", "Portugal", "Slovak Republic", "Slovenia", "Spain", "Sweden", "Switzerland", "Turkey", "United Kingdom", "United States"), class = "factor"), X2005 = structure(c(26L, 2L, 34L, 33L, 13L, 14L), .Label = c("10,3533", "10,4187", "10,8089", "10,8629", "10,882", "11,0173", "15,8399", "5,0226", "5,4488", "5,6273", "5,8713", "6,2137", "6,6397", "6,9339", "7,0448", "7,5719", "7,8534", "7,9457", "8,1819", "8,2668", "8,2883", "8,3556", "8,394", "8,4295", "8,4456", "8,4794", "8,7437", "9,0304", "9,0615", "9,4427", "9,6618", "9,77", "9,8295", "9,9833"), class = "factor"), X2006 = structure(c(25L, 2L, 31L, 34L, 13L, 14L), .Label = c("10,0326", "10,2177", "10,3877", "10,6374", "10,7468", "10,9516", "15,9368", "5,0169", "5,6845", "5,8109", "6,1019", "6,2008", "6,285", "6,6937", "7,3477", "7,5148", "7,5836", "7,7495", "8,1986", "8,2586", "8,2807", "8,3448", "8,39", "8,4289", "8,5204", "8,564", "8,8247", "8,8401", "8,948", "9,1292", "9,4811", "9,7487", "9,9243", "9,9621"), class = "factor"), X2007 = structure(c(27L, 3L, 31L, 1L, 14L, 13L), .Label = c("10,0263", "10,2099", "10,2617", "10,4771", "10,7642", "10,8754", "16,1608", "5,1597", "5,7779", "6,0372", "6,3331", "6,3858", "6,5223", "6,5494", "7,1288", "7,6299", "7,6744", "7,7553", "7,8565", "7,9023", "8,043", "8,2295", "8,4769", "8,4908", "8,5014", "8,504", "8,5531", "8,746", "8,9172", "9,0913", "9,5254", "9,8104", "9,9873", "9,9942"), class = "factor"), X2008 = structure(c(26L, 6L, 34L, 4L, 17L, 15L), .Label = c("10,1268", "10,183", "10,2189", "10,2537", "10,289", "10,4896", "10,7042", "10,9909", "11,0232", "16,6201", "5,8474", "6,0577", "6,0745", "6,586", "6,8189", "6,8863", "7,1361", "7,1819", "7,4631", "7,7052", "8,0208", "8,3068", "8,3457", "8,5513", "8,605", "8,751", "8,8915", "8,9402", "8,9521", "9,0591", "9,1344", "9,2284", "9,3051", "9,9128"), class = "factor"), X2009 = structure(c(24L, 8L, 5L, 9L, 21L, 22L), .Label = c("", "10,0115", "10,0496", "10,1957", "10,5938", "10,8137", "11,0005", "11,1729", "11,3992", "11,4722", "11,7314", "11,7516", "11,8823", "17,6706", "6,4098", "7,039", "7,1018", "7,2127", "7,6797", "7,7356", "7,8649", "7,9514", "7,9657", "9,0423", "9,152", "9,17", "9,1947", "9,4037", "9,5258", "9,6247", "9,636", "9,6743", "9,9056", "9,939"), class = "factor"), X2010 = structure(c(23L, 6L, 3L, 8L, 18L, 19L), .Label = c("", "10,1995", "10,503", "10,797", "10,8817", "11,0318", "11,0751", "11,3738", "11,5495", "11,677", "12,0661", "17,6911", "6,1782", "6,3394", "7,0229", "7,1675", "7,2911", "7,37", "7,4319", "7,6856", "8,0302", "8,8718", "8,9481", "8,9888", "8,995", "9,2925", "9,312", "9,4079", "9,4224", "9,4688", "9,5277", "9,5504", "9,589", "9,6074"), class = "factor"), X2011 = structure(c(NA, 5L, 4L, 8L, 18L, 17L), .Label = c("", "10,2345", "10,2844", "10,5139", "10,7769", "10,8671", "11,0148", "11,1784", "11,3323", "11,6343", "11,9369", "17,683", "5,922", "6,6464", "6,8709", "7,3692", "7,5011", "7,5206", "7,7333", "7,8877", "7,9416", "8,8501", "8,9042", "9,0019", "9,027", "9,1296", "9,2256", "9,2837", "9,2969", "9,4184", "9,4661"), class = "factor"), X2012 = structure(c(NA, NA, NA, 2L, 5L, NA), .Label = c("", "11,2132", "11,2955", "7,5249", "7,6077", "7,8226", "8,7596", "8,923", "9,148", "9,167", "9,3722"), class = "factor")), .Names = c("country", "X2005", "X2006", "X2007", "X2008", "X2009", "X2010", "X2011", "X2012"), row.names = c(NA, 6L), class = "data.frame")
答案 0 :(得分：3)
您可以使用正确的格式读取数据。
例如，如果您使用read.table，请更改dec = ","参数。您可能还需要更改na.strings = "<NA>"参数。
答案 1 :(得分：2)
你可以使用for循环：
for (col in names(df[,-1])) {
df[,col] <- as.numeric(gsub(",", ".", df[,col]))
}