使用R讀取帶有兩個定界符的dat文件

Question

我正在嘗試讀取如下數據示例：

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，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，- 16 47，-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 ;：

數據分為3個塊：

1. 第一個是以“：”結尾的時間戳，我們可以將其保留為數字
2. 然后是多組數字（六個），以“;：”結尾
3. 最后是第三個塊（7個元素，在字符串和數字之間混合），以“;：”結尾

是否有一種優雅的方法將此數據讀入R數據幀？ 我試過了

read.table("855360.dat",
                        header = FALSE,
                            sep = ";") 

但是需要大量的操作才能將元素設置為3個大塊，我可以將它們加入並操縱？

Answer 1

如果單個數據幀結果正常，則只需用逗號替換冒號和分號，然后將其讀取：

L <- readLines("myfile")
read.table(text =  gsub("[:;]+", ",", L), sep = ",", as.is = TRUE)

或者，如果您想生成一個嵌套列表結構，則從上方使用L ：

lapply(lapply(strsplit(L, ":"), strsplit, ";"), lapply, strsplit, ",")

Answer 2

如果將其轉換為多遍處理，則可能會花費更長的時間，但從長遠來看，可能更易於修改和維護。

如果先將字符串用";:"字符分隔，您會注意到奇數索引是主要數據（包括時間戳記），偶數索引是您的“第三塊”，其中包含混合的num / char項。 分解后，您可能會意識到我們仍然有一個解析問題，但這要簡單一些。

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"

這里的一個重要假設是，我們將始終有一對時間戳/數據和后續塊：

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"

從這里開始，意識到我們可以輕松地用另一個strsplit提取時間戳，然后可以通過替換";"將其余部分轉換成類似read.csv ";" 與換行符（與第三塊相同）：

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))

看一下其中一對數據：

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

這是一個很好的嵌套列表數據描述。 我故意將timestamp為data.frame以簡化以后的步驟，盡管這當然不是data.frame 。

如果要在包含所有數據的單個data.frame進行描述， data.frame兩件事：

1. 您的timestamp和“第三塊數據”將在數據中的所有行中重復。 根據您打算如何使用數據，這可能不是問題。 如果“第三塊”中單行數據的假設無效，則此方法會中斷 。
2. 我們在兩個數據元素中都有相同的列名。 如果您有預定義的列（總是6和7列），或者如果在數據中定義了列（在您的示例中沒有），則很容易避免此問題。 如果這些都不起作用，那么您需要確定一個適合您的命名約定。 為了這個示例，我將第二個data.frame從V1命名更改為X1命名。

牢記2：

data2mod <- lapply(data2, function(df) setNames(df, paste("X", seq_along(df), sep = "")))
bothlst2 <- mapply(list, timestamps, data1, data2mod, SIMPLIFY = FALSE)

現在，對於每個元素，我們可以將元素“列綁定”到單個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 ...

從這里開始，直接處理它們或以類似的方式將它們組合起來是很直接的：

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

基於上面的第一個項目符號，您會注意到timestamp列和所有X*列都是多余的，類似於表的聯接。