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I am trying to recode variables, which lie in columns of multiple dataframes. These dataframes also lie within lists. I have an attempt using a loop which works. I wonder if there is an even easier possibility using lapply.
My lapply attempt did not work, because the recode command tried accessing the dataframe, but it needs the columns. Is there an easy lapply way to do it?
This is an example of how my data (dataframes within list) looks like using the structure command str():
> str(Q1_dfcl)
List of 19
$ Q1_Ques_01Dep:'data.frame': 384 obs. of 20 variables:
..$ Q001_01: num [1:384] 3 2 2 1 4 2 1 1 3 1 ...
..$ Q001_02: num [1:384] 2 1 3 1 5 1 1 1 1 1 ...
..$ Q001_05: num [1:384] 3 1 3 1 2 2 2 1 1 1 ...
..$ Q001_06: num [1:384] 2 2 5 2 5 1 2 1 1 1 ...
..$ Q001_08: num [1:384] 3 1 5 1 4 1 1 1 1 1 ...
..$ Q001_09: num [1:384] 3 1 2 2 3 3 1 2 3 1 ...
..$ Q001_11: num [1:384] 4 2 4 1 1 3 1 1 2 2 ...
..$ Q001_13: num [1:384] 1 1 3 1 2 1 1 1 1 1 ...
..$ Q001_21: num [1:384] 3 1 5 3 5 1 2 1 2 1 ...
..$ Q001_26: num [1:384] 3 2 5 1 4 2 1 1 1 1 ...
..$ Q001_27: num [1:384] 4 1 4 1 5 2 2 2 1 2 ...
..$ Q001_30: num [1:384] 2 3 5 2 4 1 1 2 1 1 ...
..$ Q001_31: num [1:384] 3 1 5 2 5 1 1 1 1 2 ...
..$ Q001_40: num [1:384] 2 1 5 2 5 1 2 1 1 1 ...
..$ Q001_48: num [1:384] 4 1 5 1 4 2 2 1 2 1 ...
..$ Q001_51: num [1:384] 3 1 5 3 4 1 1 1 2 1 ...
..$ Q001_52: num [1:384] 1 1 2 1 2 1 1 1 1 1 ...
..$ Q001_57: num [1:384] 1 1 1 1 1 1 1 1 1 1 ...
..$ Q001_61: num [1:384] 2 2 2 1 2 2 1 1 1 1 ...
..$ Q001_64: num [1:384] 4 4 5 3 5 3 2 2 5 3 ...
$ Q1_Ques_02Dys:'data.frame': 384 obs. of 10 variables:
..$ Q001_02: num [1:384] 2 1 3 1 5 1 1 1 1 1 ...
..$ Q001_05: num [1:384] 3 1 3 1 2 2 2 1 1 1 ...
..$ Q001_08: num [1:384] 3 1 5 1 4 1 1 1 1 1 ...
..$ Q001_09: num [1:384] 3 1 2 2 3 3 1 2 3 1 ...
..$ Q001_21: num [1:384] 3 1 5 3 5 1 2 1 2 1 ...
..$ Q001_31: num [1:384] 3 1 5 2 5 1 1 1 1 2 ...
..$ Q001_40: num [1:384] 2 1 5 2 5 1 2 1 1 1 ...
..$ Q001_48: num [1:384] 4 1 5 1 4 2 2 1 2 1 ...
..$ Q001_57: num [1:384] 1 1 1 1 1 1 1 1 1 1 ...
..$ Q001_61: num [1:384] 2 2 2 1 2 2 1 1 1 1 ...
$ Q1_Ques_03Las:'data.frame': 384 obs. of 6 variables:
..$ Q001_06: num [1:384] 2 2 5 2 5 1 2 1 1 1 ...
..$ Q001_29: num [1:384] 3 1 2 2 5 1 1 1 1 1 ...
..$ Q001_30: num [1:384] 2 3 5 2 4 1 1 2 1 1 ...
..$ Q001_43: num [1:384] 3 2 2 2 1 2 1 1 1 1 ...
..$ Q001_54: num [1:384] 1 2 3 1 4 1 1 1 1 2 ...
..$ Q001_55: num [1:384] 2 2 5 1 5 1 1 1 3 1 ...
This is what my code looks like:
### Preparing the values for for recoding
level_key_difficulty <- c("1" = "0", "2" = "1", "3" = "2", "4" = "3", "5" = "4")
### This works.
for (i in 1:length(Q1_list)){
for (j in 1:length(Q1_list[[i]])){
Q1_dfcl[[i]][j] <- lapply(Q1_dfcl[[i]][j], function(x) recode(x, !!!level_key_difficulty))[[1]]
names(Q1_dfcl[[i]][j]) <- names(Q1_list[[i]][j])
}
}
### This does not work, because it only tries to do it on the WHOLE dataframe within
# Q1_dfcl[[1]]
lapply(Q1_dfcl, \(x) recode(x, !!!level_key_difficulty))
Actually you could just subtract 1.
dlst ## before
# [[1]]
# V1 V2 V3 V4
# 1 1 1 2 4
# 2 5 2 2 1
# 3 1 4 1 5
#
# [[2]]
# V1 V2 V3 V4
# 1 4 3 3 5
# 2 2 1 4 5
# 3 2 1 5 4
dlst_new <- lapply(dlst, `-`, 1)
dlst_new ## after
# [[1]]
# V1 V2 V3 V4
# 1 0 0 1 3
# 2 4 1 1 0
# 3 0 3 0 4
#
# [[2]]
# V1 V2 V3 V4
# 1 3 2 2 4
# 2 1 0 3 4
# 3 1 0 4 3
To change just the values of specific columns, we can do
sset <- c('V1', 'V2') ## define subset
dlst_new1 <- lapply(dlst, \(x) {x[sset] <- x[sset] - 1; x})
dlst_new1
# [[1]]
# V1 V2 V3 V4
# 1 0 0 2 4
# 2 4 1 2 1
# 3 0 3 1 5
#
# [[2]]
# V1 V2 V3 V4
# 1 3 2 3 5
# 2 1 0 4 5
# 3 1 0 5 4
Data:
dlst <- list(structure(list(V1 = c(1L, 5L, 1L), V2 = c(1L, 2L, 4L), V3 = c(2L,
2L, 1L), V4 = c(4L, 1L, 5L)), class = "data.frame", row.names = c(NA,
-3L)), structure(list(V1 = c(4L, 2L, 2L), V2 = c(3L, 1L, 1L),
V3 = 3:5, V4 = c(5L, 5L, 4L)), class = "data.frame", row.names = c(NA,
-3L)))
It can be done when you nest your lapply such as:
lapply(list(mtcars), \(a) lapply(a, \(b) recode(b, !!!level_key_difficulty, .default = "def")))
Purrr has a convenient function called map_depth that allows you to map over a list of many depths, in this case you want to go to depth two since the first depth is just a data.frame. The second depth is the columns of interest
purrr::map_depth(list(mtcars), 2, ~ recode(.x, !!!level_key_difficulty, .default = "def"))
There is also rapply a function I rarely use but is similar to map_depth
rapply(list(mtcars), \(a) recode(a, !!!level_key_difficulty, .default = "def"), how = "list")
I face problem in r, while doing glm. The problem is, Variable length differ found for "var1". but when I delete this var1 from the data. Then next same type error appear for next variables present in the data. I checked all the data, but there are no length differs in actual. How I resolve this problem? Anyone can please help me. Thanks in advance.
The data is look like; d_status is my response variable and is factor. here doesn't appear because of there are more variables.
data.frame': 300 obs. of 20 variables:
$ age : num 28 43 32 64 37 42 36 48 55 31 ...
$ gender : num 1 2 2 2 1 2 2 1 2 2 ...
$ u_clarity: num 1 2 1 2 1 1 1 2 1 1 ...
$ ph : num 5 5.5 5 5 5 5 5 5.2 5 5 ...
$ sp_g : num 1.01 1.02 1.01 1.01 1.01 ...
$ albumin : num 1 1 2 1 2 2 2 1 1 2 ...
$ glucose : num 2 2 2 2 2 2 2 2 2 2 ...
$ sugar : num 2 1 2 2 2 2 1 1 1 2 ...
$ kb : num 2 2 2 2 2 2 2 2 2 2 ...
$ bpigment : num 2 2 2 2 2 2 2 1 2 2 ...
$ ur_bi : num 2 2 2 2 2 2 2 2 2 2 ...
$ blood : num 2 2 2 2 2 2 2 2 2 2 ...
$ pus_cells: num 1 2 1 2 1 1 1 1 1 1 ...
$ red_cells: num 1 2 1 2 1 1 1 2 1 1 ...
$ epi_cells: num 1 2 1 2 1 2 1 1 2 2 ...
$ mt : num 1 2 1 2 1 1 2 2 2 1 ...
$ co : num 2 1 1 2 1 1 1 2 2 1 ...
$ gc : num 2 1 1 1 1 1 1 2 2 1 ...
$ bacteria : num 1 2 1 2 1 1 1 2 2 1 ...
$ cc : num 1 1 1 2 1 1 1 2 1 1 ...
f1=glm(y~.,family=quasibinomial(link='logit'),data=dataset1[training,])
Error in model.frame.default(formula = y ~ ., data = dataset1[training, :
variable lengths differ (found for 'age')
I have run a logistical regression model with both categorical and numerical variables. The model was trying to predict the number of website visits in a month based off the first week. Obviously the number of website visits in the first week was the strongest indicator. However when i ran h2o deep learning with various models and activation functions the model performs very poorly. Based off the var_imp function it gives importance to very non important variables(based off my logistical regression model, which is quite good, this is wrong), and only seems to have categorical subsets ranked with high importance. and the model does not perform well even on the training data, a real warning sign! So i just wanted to upload my code to check i am not doing anything to harm the model. It seems strange for logistical regression to get it quiteright but deep learning to get it so wrong, so i imagine its something i've done!
summary(data)
$ VAR1: Factor w/ 8 levels ,..: 1 5 2 1 7 2 5 1 5 1 ...
$ VAR2: Factor w/ 5 levels ,..: 1 4 1 1 4 4 4 1 1 4 ...
$ VAR3: Factor w/ 2 levels "F","M": 2 2 2 1 2 2 2 2 2 2 ...
$ VAR4: Factor w/ 2 levels : 2 1 2 2 1 1 1 2 2 1 ...
$ VAR5 : num 1000 20 30 20 30 30 30 50 30 400 ...
$ VAR6: Factor w/ 2 levels "N","Y": 1 2 2 1 2 2 2 2 1 2 ...
$ VAR7: Factor w/ 2 levels "N","Y": 1 2 2 1 2 2 2 2 1 2 ...
$ VAR8: num 0 0 0 0 0 0 0 0 0 0 ...
$ VAR9: num 56 52 49 29 28 38 34 79 53 36 ...
$ VAR10: num 3 2 1 3 2 2 3 4 2 2 ...
$ VAR11: num 1 1 1 2 2 1 1 1 1 2 ...
$ VAR12: Factor w/ 2 levels "N","Y": 1 1 1 1 2 1 1 1 1 1 ...
$ VAR13: num 1 0 1 1 1 0 1 0 0 0 ...
$ VAR14: Factor w/ 2 levels "N","Y": 2 1 1 1 1 1 1 1 1 1 ...
$ VAR15: Factor w/ 2 levels "N","Y": 1 1 1 1 1 1 1 1 1 1 ...
$ VAR16: num 1 0 0 1 0 0 0 1 1 0 ...
$ VAR17: num 19 7 1 4 10 2 4 4 7 12 ...
$ VAR18: Factor w/ 2 levels "N","Y": 1 2 2 2 2 2 2 1 2 1 ...
$ VAR19: Factor w/ 2 levels "0","Y": 1 1 2 1 1 1 1 1 1 1 ...
$ VAR20: Factor w/ 2 levels "N","Y": 1 1 2 1 1 1 1 1 1 1 ...
$ VAR21: Factor w/ 2 levels "N","Y": 1 1 1 1 1 1 1 1 1 1 ...
$ VAR22: : num 0.579 0 0 0 0.4 ...
$ VAR23: num 1.89 1 1 1 2.9 ...
$ VAR24: num 0.02962 0.00691 0.05327 0.02727 0.01043 ...
$ VAR25: Factor w/ 3 levels ..: 2 2 2 3 3 2 3 2 1 3 ...
$ VAR26: num 3 2 1 2 3 1 2 1 2 4 ...
$ VAR27: num 3 2 1 1 5 1 1 1 1 2 ...
$ VAR_RESPONSE: num 7 24 4 3 8 12 5 48 2 7 ...
sapply(data,function(x) sum(is.na(x)))
colSums(is.na(data))
data[is.na(data)] = 0
d.hex = as.h2o(data, destination_frame= "d.hex")
Data_g.split = h2o.splitFrame(data = d.hex,ratios = 0.75)
Data_train = Data_g.split[[1]]#75% training data
Data_test = Data_g.split[[2]]
activation_opt <-
c("Rectifier","RectifierWithDropout","Maxout","MaxoutWithDropout",
"Tanh","TanhWithDropout")
hidden_opt <- list(c(10,10),c(20,15),c(50,50,50),c(5,3,2),c(100,100),c(5),c(30,30,30),c(50,50,50,50),c(5,4,3,2))
l1_opt <- c(0,1e-3,1e-5,1e-7,1e-9)
l2_opt <- c(0,1e-3,1e-5,1e-7,1e-9)
hyper_params <- list( activation=activation_opt,
hidden=hidden_opt,
l1=l1_opt,
l2=l2_opt )
search_criteria <- list(strategy = "RandomDiscrete", max_models=30)
dl_grid10 <- h2o.grid("deeplearning"
,grid_id = "deep_learn10"
,hyper_params = hyper_params
,search_criteria = search_criteria
,x = 1:27
,y = "VAR_RESPONSE"
,training_frame = Data_train)
d_grid10 <- h2o.getGrid("deep_learn10",sort_by = "mse")
mn = h2o.deeplearning(x = 1:27,
y = "VAR_RESPONSE",
training_frame = Data_train,
model_id = "mn",
activation = "Maxout",
l1 = 0,
l2 = 1e-9,
hidden = c(100,100),)
how could I write a list of matrices into a csv file in R?
I tried
fff = tr[[3]]
MATweight = ldply(fff, function(t) t$toDataFrame())
but had this error
Error in t$toDataFrame : $ operator is invalid for atomic vectors
I am not sure if it was the right thing to do, any idea please?
I have this list of matrices
> str(fff)
List of 10
$ : num [1:1000, 1:50] 1 1 1 1 2 2 1 2 2 1 ...
....
$ : num [1:1000, 1:50] 1 1 1 1 2 2 1 2 2 1 ..
.
When I tried the suggested answer by G. I got:
> write.csv(map_dfr(fff, as.data.frame, .id = "matrix"),"testt.csv", row.names = FALSE)
> tgtg = read.csv("/Users/amani/testt.csv")
> str(tgtg)
'data.frame': 10000 obs. of 51 variables:
$ matrix: int 1 1 1 1 1 1 1 1 1 1 ...
$ V1 : int 1 1 1 1 2 2 1 2 2 1 ...
$ V2 : int 2 2 1 2 1 1 2 1 1 2 ...
$ V3 : int 1 1 1 1 1 2 1 1 2 1 ...
....
$ V50 : int 2 2 1 1 1 1 1 1 2 2 ...
Using the L test data shown, try map_dfr from purrr:
library(purrr)
L <- list(as.matrix(BOD), as.matrix(10*BOD)) # test data
write.csv(map_dfr(L, as.data.frame, .id = "matrix"), stdout(), row.names = FALSE)
giving:
"matrix","Time","demand"
"1",1,8.3
"1",2,10.3
"1",3,19
"1",4,16
"1",5,15.6
"1",7,19.8
"2",10,83
"2",20,103
"2",30,190
"2",40,160
"2",50,156
"2",70,198
Why not do it like this:
# sample data like your `fff` list of matrices
fff <- lapply(1:10, function(i)
matrix(data = sample(1:2, 510000, replace = TRUE), nrow = 10000, ncol = 51))
# we give each matrix in the list a unique name
mat_names <- as.character(1:length(fff))
# converting to dataframe with the list index preserved
fffdf <- lapply(1:length(fff), function(i)
cbind(mat_names = mat_names[i], as.data.frame(fff[[i]])))
fffdf <- do.call(rbind, fffdf)
# writing it to a file
t <- tempfile(fileext = ".csv")
write.csv(fffdf, t, row.names = FALSE)
# testing the reassembly into the same list of matrices that you started
tcsv <- read.csv(t)
tgtg <- lapply(mat_names, function(i) {
df <- subset(tcsv, mat_names == i, select = -mat_names)
mat <- as.matrix(df)
dimnames(mat) <- NULL
return(mat)
})
Testing the result: is tgtg the same as the fff you began with?
> identical(fff, tgtg)
[1] TRUE
> str(tgtg)
List of 10
$ : int [1:10000, 1:51] 1 2 1 2 1 1 2 2 1 1 ...
$ : int [1:10000, 1:51] 2 2 2 2 2 2 1 1 1 1 ...
$ : int [1:10000, 1:51] 1 2 1 1 2 2 1 1 2 1 ...
$ : int [1:10000, 1:51] 1 1 2 2 1 1 1 2 1 1 ...
$ : int [1:10000, 1:51] 1 1 2 1 2 2 1 1 1 2 ...
$ : int [1:10000, 1:51] 2 1 2 2 1 1 1 2 1 2 ...
$ : int [1:10000, 1:51] 2 1 2 1 2 1 1 1 2 1 ...
$ : int [1:10000, 1:51] 2 2 1 1 2 2 1 2 1 1 ...
$ : int [1:10000, 1:51] 1 2 1 1 1 1 1 2 2 1 ...
$ : int [1:10000, 1:51] 1 1 1 2 2 2 2 2 2 2 ...
I'm trying to convert a triple nested list into a dataframe. This question has helped, but I can't get the dataframe I'd like.
The list is an options chain obtained from IBrokers, a summary is shown below. I've uploaded the actual chain here which is more detailed.
Chain <-
list(
list(
list(
list(version="8",contract=list(symbol="BHP",right="C",expiry="20180621",strike="25")),
list(version="8",contract=list(symbol="BHP",right="C",expiry="20180621",strike="26"))
),
list(
list(version="8",contract=list(symbol="BHP",right="C",expiry="20180730",strike="25")),
list(version="8",contract=list(symbol="BHP",right="C",expiry="20180730",strike="26"))
)
),
list(
list(
list(version="8",contract=list(symbol="CBA",right="C",expiry="20180621",strike="65")),
list(version="8",contract=list(symbol="CBA",right="C",expiry="20180621",strike="64"))
),
list(
list(version="8",contract=list(symbol="CBA",right="C",expiry="20180730",strike="65")),
list(version="8",contract=list(symbol="CBA",right="C",expiry="20180730",strike="64"))
)
)
)
I'd like to convert the list into a dataframe like this:
Contracts <- data.frame(symbol=c("BHP","BHP","BHP","BHP","CBA","CBA","CBA","CBA"),
right=c("C","C","C","C","C","C","C","C"),
expiry=c("20180621","20180621","20180730","20180730","20180621","20180621","20180730","20180730"),
strike=c("25","26","25","26","65","64","65","64"))
I tried this code, but it didn't give me the dataframe I wanted.
X <- lapply(Chain,function(x) as.data.frame.list(lapply(x,as.data.frame.list)))
dfx <- do.call(rbind,X)
Any suggestions please?
How about the following?
df <- as.data.frame(matrix(unlist(Chain, recursive = T), ncol = 5, byrow = T)[, -1]);
colnames(df) <- c("symbol", "right", "expiry", "strike");
# symbol right expiry strike
#1 BHP C 20180621 25
#2 BHP C 20180621 26
#3 BHP C 20180730 25
#4 BHP C 20180730 26
#5 CBA C 20180621 65
#6 CBA C 20180621 64
#7 CBA C 20180730 65
#8 CBA C 20180730 64
Explanation: Recursively unlist the nested Chain, then recast as matrix, remove column version and convert to data.frame. The only minor down-side is that we have to manually add column names.
Update
Since your actual data is quite different, here is a possibility.
Note: I assume the structure from the Gist is stored in tbl.
tbl;
#Source: local data frame [2 x 6]
#Groups: <by row>
#
## A tibble: 2 x 6
# symbol sectype exch currency multiplier Chain
# <fct> <fct> <fct> <fct> <fct> <list>
#1 BHP OPT ASX AUD 100 <list [1,241]>
#2 CBA OPT ASX AUD 100 <list [1,204]>
The following list contains two data.frames, one for each row from tbl.
lst <- lapply(tbl$Chain, function(x)
do.call(rbind.data.frame, lapply(x, function(y) as.data.frame(unclass(y$contract)))))
#List of 2
# $ :'data.frame': 1241 obs. of 16 variables:
# ..$ conId : Factor w/ 1241 levels "198440202","198440207",..: 1 2 3 4 5 6 7 8 9 10 ...
# ..$ symbol : Factor w/ 1 level "BHP": 1 1 1 1 1 1 1 1 1 1 ...
# ..$ sectype : Factor w/ 1 level "OPT": 1 1 1 1 1 1 1 1 1 1 ...
# ..$ exch : Factor w/ 1 level "ASX": 1 1 1 1 1 1 1 1 1 1 ...
# ..$ primary : Factor w/ 1 level "": 1 1 1 1 1 1 1 1 1 1 ...
# ..$ expiry : Factor w/ 18 levels "20180628","20181220",..: 1 1 1 1 1 1 1 1 1 1 ...
# ..$ strike : Factor w/ 118 levels "25","26","27",..: 1 1 2 2 3 3 4 4 5 5 ...
# ..$ currency : Factor w/ 1 level "AUD": 1 1 1 1 1 1 1 1 1 1 ...
# ..$ right : Factor w/ 2 levels "C","P": 1 2 1 2 1 2 1 2 1 2 ...
# ..$ local : Factor w/ 1241 levels "BHPV78","BHPV88",..: 1 2 3 4 5 6 7 8 9 10 ...
# ..$ multiplier : Factor w/ 1 level "100": 1 1 1 1 1 1 1 1 1 1 ...
# ..$ combo_legs_desc: Factor w/ 1 level "": 1 1 1 1 1 1 1 1 1 1 ...
# ..$ comboleg : Factor w/ 1 level "": 1 1 1 1 1 1 1 1 1 1 ...
# ..$ include_expired: Factor w/ 1 level "": 1 1 1 1 1 1 1 1 1 1 ...
# ..$ secIdType : Factor w/ 1 level "": 1 1 1 1 1 1 1 1 1 1 ...
# ..$ secId : Factor w/ 1 level "": 1 1 1 1 1 1 1 1 1 1 ...
# $ :'data.frame': 1204 obs. of 16 variables:
# ..$ conId : Factor w/ 1204 levels "198447027","198447030",..: 1 2 3 4 5 6 7 8 9 10 ...
# ..$ symbol : Factor w/ 1 level "CBA": 1 1 1 1 1 1 1 1 1 1 ...
# ..$ sectype : Factor w/ 1 level "OPT": 1 1 1 1 1 1 1 1 1 1 ...
# ..$ exch : Factor w/ 1 level "ASX": 1 1 1 1 1 1 1 1 1 1 ...
# ..$ primary : Factor w/ 1 level "": 1 1 1 1 1 1 1 1 1 1 ...
# ..$ expiry : Factor w/ 18 levels "20180628","20181220",..: 1 1 1 1 1 1 1 1 1 1 ...
# ..$ strike : Factor w/ 179 levels "79.68","81.68",..: 1 1 2 2 3 3 4 4 5 5 ...
# ..$ currency : Factor w/ 1 level "AUD": 1 1 1 1 1 1 1 1 1 1 ...
# ..$ right : Factor w/ 2 levels "C","P": 1 2 1 2 1 2 1 2 1 2 ...
# ..$ local : Factor w/ 1204 levels "CBAKT9","CBAKU9",..: 1 2 3 4 5 6 7 8 9 10 ...
# ..$ multiplier : Factor w/ 1 level "100": 1 1 1 1 1 1 1 1 1 1 ...
# ..$ combo_legs_desc: Factor w/ 1 level "": 1 1 1 1 1 1 1 1 1 1 ...
# ..$ comboleg : Factor w/ 1 level "": 1 1 1 1 1 1 1 1 1 1 ...
# ..$ include_expired: Factor w/ 1 level "": 1 1 1 1 1 1 1 1 1 1 ...
# ..$ secIdType : Factor w/ 1 level "": 1 1 1 1 1 1 1 1 1 1 ...
# ..$ secId : Factor w/ 1 level "": 1 1 1 1 1 1 1 1 1 1 ...
You can use unstack
unstack(data.frame(d<-unlist(Chain),names(d)))
contract.expiry contract.right contract.strike contract.symbol version
1 20180621 C 25 BHP 8
2 20180621 C 26 BHP 8
3 20180730 C 25 BHP 8
4 20180730 C 26 BHP 8
5 20180621 C 65 CBA 8
6 20180621 C 64 CBA 8
7 20180730 C 65 CBA 8
8 20180730 C 64 CBA 8
If you want you can delete the word contract.
unstack(data.frame(d<-unlist(Chain),sub(".*[.]","",names(d))))
expiry right strike symbol version
1 20180621 C 25 BHP 8
2 20180621 C 26 BHP 8
3 20180730 C 25 BHP 8
4 20180730 C 26 BHP 8
5 20180621 C 65 CBA 8
6 20180621 C 64 CBA 8
7 20180730 C 65 CBA 8
8 20180730 C 64 CBA 8
This can also be written as unstack(data.frame(d<-unlist(Chain),sub("contract[.]","",names(d)))) Although I would prefer to maintain the name contract in order to know which columns indeed form the contract dataframe needed
Or even you can change the names After unstacking.
With the new data:
a=readLines("https://raw.githubusercontent.com/hughandersen/OptionsTrading/master/Stocks_option_chain")
b=eval(parse(text=paste(a,collapse="")))
s=unstack(data.frame(d<-unlist(b[6]),names(d)))