我正在尝试阅读下面的数据示例：
1344428：-1,1，-1,415，-649,0.00; -1,2，-1,1090，-2167,0.00; -1,3，-1，-881，-3164,0.00; - 1,4，-1，-624,1529,0.00; -1,5-，-1，-849，-2875,0.00; 1,6，-1,856，-2341,0.00; -1,7，-1,758 ，-2408,0.00; -1,8-，-1，-201，-2307,0.00; -1,9-，-1，-963，-2807,0.00; -1,10，-1，-460， - 3309,0.00; -1,11，-1，-1645，-1773,0.00; -1,12，-1,1487，-518,0.00; -1,13，-1,685，-3113,0.00; -1 ，14，-1，-935，-3217,0.00; -1,15，-1，-1101，-2430,0.00; -1,16，-1,754，-2946,0.00; -1,17，-1,823 ，-2497,0.00; -1,18，-1，-948，-2431,0.00; -1,19，-1,774，-2242,0.00; -1,20，-1,861，-2192,0.00; -1 ，21，-1,433，-3391,0.00; -1,22，-1,133，-2190,0.00; -1,23，-1，-977，-2585,0.00; -1,24，-1，-968 ，-2107,0.00; -1,25，-1,175，-3062,0.00; -1,26，-1,265，-2736,0.00; -1,27，-1,67，-2735,0.00; -1， 28，-1，-281，-2752,0.00; 4,29，-1,5550,4400,0.00;：174，-2563,11,28.67，A，死，SetAway ;: 1344429：-1,1，-1,415，-649,0.00; -1,2，-1,1090，-2167,0.00; -1,3，-1，-885，-3169,0.00; -1,4- ，-1，-626,1527,0.00; -1,5-，-1，-852，-2887,0.00; 1,6，-1,854，-2340,0.00; -1,7，-1,761，-2411 ，0.00; -1,8-，-1，-201，-2307,0.00; -1,9-，-1，-967，-2808,0.00; -1,10，-1，-460，-3309,0.00 ; -1,11，-1，-1647，-1777,0.00; -1,12，-1,1485，-518,0.00; -1,13，-1,687，-3118,0.00; -1,14， -1，-938，-3222,0.00; -1,15，-1，-1100，-2430,0.00; -1,16，-1,744，-2946,0.00; -1,17，-1,815，-2505 ，0.00; -1,18，-1，-950，-2429,0.00; -1,19，-1,773，-2237,0.00; -1,20，-1,861，-2190,0.00; -1,21， -1,433，-3392,0.00; -1,22，-1,133，-2189,0.00; -1,23，-1，-980，-2593,0.00; -1,24，-1，-961，-2109 ，0.00; -1,25，-1,176，-3056,0.00; -1,26，-1,265，-2731,0.00; -1,27，-1,67，-2736,0.00; -1,28， - 1，-283，-2746,0.00; 4,29，-1,5550,4400,0.00;：174，-2563,11,28.67，A，死，SetAway;：
数据分为3个块：
有没有一种优雅的方法将这些数据读入R数据框？我试过了
read.table("855360.dat",
header = FALSE,
sep = ";")
但它需要大量的操作才能将元素设置为3个块，我可以将它们连接和操作？
答案 0 :(得分：2)
如果单个数据框结果正常，则只需用逗号替换冒号和分号，然后将其读入：
L <- readLines("myfile")
read.table(text = gsub("[:;]+", ",", L), sep = ",", as.is = TRUE)
或者如果要生成嵌套列表结构，请使用上面的L：
lapply(lapply(strsplit(L, ":"), strsplit, ";"), lapply, strsplit, ",")
答案 1 :(得分：0)
如果将其转换为多遍过程，可能需要更长的时间，但从长远来看可能更容易修改和维护。
如果你首先用";:"字符拆分字符串，你会注意到奇数索引是主要数据（包括时间戳），偶数索引是你的＆＃34;第三块＆＃34;混合num / char条目。一旦你打破了这个，你可能会发现我们仍然有一个解析问题，但它有点简单。
txt <- "1344428:-1,1,-1,415,-649,0.00;-1,2,-1,1090,-2167,0.00;-1,3,-1,-881,-3164,0.00;-1,4,-1,-624,1529,0.00;-1,5,-1,-849,-2875,0.00;-1,6,-1,856,-2341,0.00;-1,7,-1,758,-2408,0.00;-1,8,-1,-201,-2307,0.00;-1,9,-1,-963,-2807,0.00;-1,10,-1,-460,-3309,0.00;-1,11,-1,-1645,-1773,0.00;-1,12,-1,1487,-518,0.00;-1,13,-1,685,-3113,0.00;-1,14,-1,-935,-3217,0.00;-1,15,-1,-1101,-2430,0.00;-1,16,-1,754,-2946,0.00;-1,17,-1,823,-2497,0.00;-1,18,-1,-948,-2431,0.00;-1,19,-1,774,-2242,0.00;-1,20,-1,861,-2192,0.00;-1,21,-1,433,-3391,0.00;-1,22,-1,133,-2190,0.00;-1,23,-1,-977,-2585,0.00;-1,24,-1,-968,-2107,0.00;-1,25,-1,175,-3062,0.00;-1,26,-1,265,-2736,0.00;-1,27,-1,67,-2735,0.00;-1,28,-1,-281,-2752,0.00;4,29,-1,5550,4400,0.00;:174,-2563,11,28.67,A,Dead,SetAway;: 1344429:-1,1,-1,415,-649,0.00;-1,2,-1,1090,-2167,0.00;-1,3,-1,-885,-3169,0.00;-1,4,-1,-626,1527,0.00;-1,5,-1,-852,-2887,0.00;-1,6,-1,854,-2340,0.00;-1,7,-1,761,-2411,0.00;-1,8,-1,-201,-2307,0.00;-1,9,-1,-967,-2808,0.00;-1,10,-1,-460,-3309,0.00;-1,11,-1,-1647,-1777,0.00;-1,12,-1,1485,-518,0.00;-1,13,-1,687,-3118,0.00;-1,14,-1,-938,-3222,0.00;-1,15,-1,-1100,-2430,0.00;-1,16,-1,744,-2946,0.00;-1,17,-1,815,-2505,0.00;-1,18,-1,-950,-2429,0.00;-1,19,-1,773,-2237,0.00;-1,20,-1,861,-2190,0.00;-1,21,-1,433,-3392,0.00;-1,22,-1,133,-2189,0.00;-1,23,-1,-980,-2593,0.00;-1,24,-1,-961,-2109,0.00;-1,25,-1,176,-3056,0.00;-1,26,-1,265,-2731,0.00;-1,27,-1,67,-2736,0.00;-1,28,-1,-283,-2746,0.00;4,29,-1,5550,4400,0.00;:174,-2563,11,28.67,A,Dead,SetAway;:"
x <- strsplit(txt, ";:")[[1]]
x <- sapply(x, trimws, USE.NAMES = FALSE)
x[1]
# [1] "1344428:-1,1,-1,415,-649,0.00;-1,2,-1,1090,-2167,0.00;-1,3,-1,-881,-3164,0.00;-1,4,-1,-624,1529,0.00;-1,5,-1,-849,-2875,0.00;-1,6,-1,856,-2341,0.00;-1,7,-1,758,-2408,0.00;-1,8,-1,-201,-2307,0.00;-1,9,-1,-963,-2807,0.00;-1,10,-1,-460,-3309,0.00;-1,11,-1,-1645,-1773,0.00;-1,12,-1,1487,-518,0.00;-1,13,-1,685,-3113,0.00;-1,14,-1,-935,-3217,0.00;-1,15,-1,-1101,-2430,0.00;-1,16,-1,754,-2946,0.00;-1,17,-1,823,-2497,0.00;-1,18,-1,-948,-2431,0.00;-1,19,-1,774,-2242,0.00;-1,20,-1,861,-2192,0.00;-1,21,-1,433,-3391,0.00;-1,22,-1,133,-2190,0.00;-1,23,-1,-977,-2585,0.00;-1,24,-1,-968,-2107,0.00;-1,25,-1,175,-3062,0.00;-1,26,-1,265,-2736,0.00;-1,27,-1,67,-2735,0.00;-1,28,-1,-281,-2752,0.00;4,29,-1,5550,4400,0.00"
x[2]
# [1] "174,-2563,11,28.67,A,Dead,SetAway"
这里一个重要的假设是我们总是会有成对的时间戳/数据和后续块：
if (length(x) %% 2 != 0) stop("oops, uneven pairs")
odds <- seq(1, length(x), by = 2)
str(x[odds])
# chr [1:2] "1344428:-1,1,-1,415,-649,0.00;-1,2,-1,1090,-2167,0.00;-1,3,-1,-881,-3164,0.00;-1,4,-1,-624,1529,0.00;-1,5,-1,-849,-2875,0.00;-1"| __truncated__ ...
x[-odds]
# [1] "174,-2563,11,28.67,A,Dead,SetAway" "174,-2563,11,28.67,A,Dead,SetAway"
从这里开始，我们意识到我们可以轻松地使用其他strsplit提取时间戳，然后通过用换行符替换read.csv来将其余内容转换为";"喜欢的内容（与您的相同）第三块）：
timestamps <- lapply(firstsplit, function(z) data.frame(timestamp = as.numeric(z[1])))
data1 <- lapply(firstsplit, function(lst) read.csv(textConnection(gsub(";", "\n", lst[[2]])), header = FALSE))
data2 <- lapply(secondsplit, function(z) read.csv(textConnection(z), header = FALSE))
查看其中一对数据：
bothlst <- mapply(list, timestamps, data1, data2, SIMPLIFY = FALSE)
str(bothlst[[1]])
# List of 3
# $ :'data.frame': 1 obs. of 1 variable:
# ..$ timestamp: num 1344428
# $ :'data.frame': 29 obs. of 6 variables:
# ..$ V1: int [1:29] -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 ...
# ..$ V2: int [1:29] 1 2 3 4 5 6 7 8 9 10 ...
# ..$ V3: int [1:29] -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 ...
# ..$ V4: int [1:29] 415 1090 -881 -624 -849 856 758 -201 -963 -460 ...
# ..$ V5: int [1:29] -649 -2167 -3164 1529 -2875 -2341 -2408 -2307 -2807 -3309 ...
# ..$ V6: num [1:29] 0 0 0 0 0 0 0 0 0 0 ...
# $ :'data.frame': 1 obs. of 7 variables:
# ..$ V1: int 174
# ..$ V2: int -2563
# ..$ V3: int 11
# ..$ V4: num 28.7
# ..$ V5: Factor w/ 1 level "A": 1
# ..$ V6: Factor w/ 1 level "Dead": 1
# ..$ V7: Factor w/ 1 level "SetAway": 1
这是一个很好的嵌套列表描述您的数据。我故意让timestamp成为data.frame以简化后续步骤，但这当然不是必需的。
如果您希望在单个data.frame中描述所有数据，请注意以下两点：
timestamp和＆＃34;第三块数据＆＃34;将在数据中的所有行上重复。这可能不是问题，具体取决于您打算如何使用数据。 如果在＆＃34;第三块＆＃34;中假设单行数据，则此方法会中断是无效的。data.frame从V1命名更改为X1命名。牢记第2条：
data2mod <- lapply(data2, function(df) setNames(df, paste("X", seq_along(df), sep = "")))
bothlst2 <- mapply(list, timestamps, data1, data2mod, SIMPLIFY = FALSE)
现在，对于每个元素，我们可以＆＃34; column-bind＆＃34;将元素合并为一个data.frame：
# bothdf <- lapply(bothlst2, cbind.data.frame)
str(bothdf)
# List of 2
# $ :'data.frame': 29 obs. of 14 variables:
# ..$ timestamp: num [1:29] 1344428 1344428 1344428 1344428 1344428 ...
# ..$ V1 : int [1:29] -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 ...
# ..$ V2 : int [1:29] 1 2 3 4 5 6 7 8 9 10 ...
# ..$ V3 : int [1:29] -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 ...
# ..$ V4 : int [1:29] 415 1090 -881 -624 -849 856 758 -201 -963 -460 ...
# ..$ V5 : int [1:29] -649 -2167 -3164 1529 -2875 -2341 -2408 -2307 -2807 -3309 ...
# ..$ V6 : num [1:29] 0 0 0 0 0 0 0 0 0 0 ...
# ..$ X1 : int [1:29] 174 174 174 174 174 174 174 174 174 174 ...
# ..$ X2 : int [1:29] -2563 -2563 -2563 -2563 -2563 -2563 -2563 -2563 -2563 -2563 ...
# ..$ X3 : int [1:29] 11 11 11 11 11 11 11 11 11 11 ...
# ..$ X4 : num [1:29] 28.7 28.7 28.7 28.7 28.7 ...
# ..$ X5 : Factor w/ 1 level "A": 1 1 1 1 1 1 1 1 1 1 ...
# ..$ X6 : Factor w/ 1 level "Dead": 1 1 1 1 1 1 1 1 1 1 ...
# ..$ X7 : Factor w/ 1 level "SetAway": 1 1 1 1 1 1 1 1 1 1 ...
# $ :'data.frame': 29 obs. of 14 variables:
# ..$ timestamp: num [1:29] 1344429 1344429 1344429 1344429 1344429 ...
# ..$ V1 : int [1:29] -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 ...
# ..$ V2 : int [1:29] 1 2 3 4 5 6 7 8 9 10 ...
# ..$ V3 : int [1:29] -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 ...
# ..$ V4 : int [1:29] 415 1090 -885 -626 -852 854 761 -201 -967 -460 ...
# ..$ V5 : int [1:29] -649 -2167 -3169 1527 -2887 -2340 -2411 -2307 -2808 -3309 ...
# ..$ V6 : num [1:29] 0 0 0 0 0 0 0 0 0 0 ...
# ..$ X1 : int [1:29] 174 174 174 174 174 174 174 174 174 174 ...
# ..$ X2 : int [1:29] -2563 -2563 -2563 -2563 -2563 -2563 -2563 -2563 -2563 -2563 ...
# ..$ X3 : int [1:29] 11 11 11 11 11 11 11 11 11 11 ...
# ..$ X4 : num [1:29] 28.7 28.7 28.7 28.7 28.7 ...
# ..$ X5 : Factor w/ 1 level "A": 1 1 1 1 1 1 1 1 1 1 ...
# ..$ X6 : Factor w/ 1 level "Dead": 1 1 1 1 1 1 1 1 1 1 ...
# ..$ X7 : Factor w/ 1 level "SetAway": 1 1 1 1 1 1 1 1 1 1 ...
从这里开始，它可以独立地处理它们，或者以类似的方式将它们组合起来：
head(do.call("rbind", bothdf))
# timestamp V1 V2 V3 V4 V5 V6 X1 X2 X3 X4 X5 X6 X7
# 1 1344428 -1 1 -1 415 -649 0 174 -2563 11 28.67 A Dead SetAway
# 2 1344428 -1 2 -1 1090 -2167 0 174 -2563 11 28.67 A Dead SetAway
# 3 1344428 -1 3 -1 -881 -3164 0 174 -2563 11 28.67 A Dead SetAway
# 4 1344428 -1 4 -1 -624 1529 0 174 -2563 11 28.67 A Dead SetAway
# 5 1344428 -1 5 -1 -849 -2875 0 174 -2563 11 28.67 A Dead SetAway
# 6 1344428 -1 6 -1 856 -2341 0 174 -2563 11 28.67 A Dead SetAway
根据我上面的第一个项目符号，您会注意到timestamp列和所有X*列都是多余的，类似于表的连接。