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how can I convert values to a meaning separated with a colon (double point)
Simultaneously merge multiple data.frames in a listCharacters counting and subletting specific patternsdata.table vs dplyr: can one do something well the other can't or does poorly?how to make a bar plot for a list of dataframes?Matplot not plotting datasetOrder of elements in CharacterVector and NumericVector in rcppFilter two tables with crosstalkR- How to use map() into map()R - Pmap() instead of Map()How to select highest value of col in r
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty
margin-bottom:0;
1
I have a data like this
df<- structure(list(df = structure(c(10L, 8L, 2L, 8L, 7L, 7L, 10L,
8L, 3L, 10L, 10L, 9L, 9L, 1L, 1L, 3L, 1L, 5L, 5L, 4L, 10L, 8L,
1L, 1L, 2L, 6L), .Label = c("-1:-1:2", "-1:2:-1", "-1:2:2", "1:01:01",
"1:1(2):1", "1(1)|1(2):1(1)|1(2):1(1)|1(2)", "1(1)|1(2):2:2",
"2:-1:-1", "2:-1:2", "2:02:02"), class = "factor")), class = "data.frame", row.names = c(NA,
-26L))
I want to expend it to words as I define. I want to have as many columns as the number of double points , for example here we have three : so we will add 3 columns after the df. Then we fill it up with words
2 = Homo
-1 = No
1= Het
1(1)= Het1
1(2)= Het2
So an expected output looks like below.
2:02:02 Homo Homo Homo
2:-1:-1 Homo No No
-1:2:-1 No Homo No
2:-1:-1 Homo No No
1(1)|1(2):2:2 Het1 Het2 Homo Homo
1(1)|1(2):2:2 Het1 Het2 Homo Homo
2:02:02 Homo Homo Homo
2:-1:-1 Homo No No
-1:2:2 No Homo Homo
2:02:02 Homo Homo Homo
2:02:02 Homo Homo Homo
2:-1:2 Homo No Homo
2:-1:2 Homo No Homo
-1:-1:2 No No Homo
-1:-1:2 No No Homo
-1:2:2 No Homo Homo
-1:-1:2 No No Homo
1:1(2):1 Het Het2 Het
1:1(2):1 Het Het3 Het
1:01:01 Het Het Het
2:02:02 Homo Homo Homo
2:-1:-1 Homo No No
-1:-1:2 No No Homo
-1:-1:2 No No Homo
-1:2:-1 No Homo No
1(1)|1(2):1(1)|1(2):1(1)|1(2) Het1 Het2 Het1 Het2 Het1 Het2
r
asked Mar 28 at 20:57
LearnerLearner
4211 silver badge11 bronze badges
add a comment
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1
I have a data like this
df<- structure(list(df = structure(c(10L, 8L, 2L, 8L, 7L, 7L, 10L,
8L, 3L, 10L, 10L, 9L, 9L, 1L, 1L, 3L, 1L, 5L, 5L, 4L, 10L, 8L,
1L, 1L, 2L, 6L), .Label = c("-1:-1:2", "-1:2:-1", "-1:2:2", "1:01:01",
"1:1(2):1", "1(1)|1(2):1(1)|1(2):1(1)|1(2)", "1(1)|1(2):2:2",
"2:-1:-1", "2:-1:2", "2:02:02"), class = "factor")), class = "data.frame", row.names = c(NA,
-26L))
I want to expend it to words as I define. I want to have as many columns as the number of double points , for example here we have three : so we will add 3 columns after the df. Then we fill it up with words
2 = Homo
-1 = No
1= Het
1(1)= Het1
1(2)= Het2
So an expected output looks like below.
2:02:02 Homo Homo Homo
2:-1:-1 Homo No No
-1:2:-1 No Homo No
2:-1:-1 Homo No No
1(1)|1(2):2:2 Het1 Het2 Homo Homo
1(1)|1(2):2:2 Het1 Het2 Homo Homo
2:02:02 Homo Homo Homo
2:-1:-1 Homo No No
-1:2:2 No Homo Homo
2:02:02 Homo Homo Homo
2:02:02 Homo Homo Homo
2:-1:2 Homo No Homo
2:-1:2 Homo No Homo
-1:-1:2 No No Homo
-1:-1:2 No No Homo
-1:2:2 No Homo Homo
-1:-1:2 No No Homo
1:1(2):1 Het Het2 Het
1:1(2):1 Het Het3 Het
1:01:01 Het Het Het
2:02:02 Homo Homo Homo
2:-1:-1 Homo No No
-1:-1:2 No No Homo
-1:-1:2 No No Homo
-1:2:-1 No Homo No
1(1)|1(2):1(1)|1(2):1(1)|1(2) Het1 Het2 Het1 Het2 Het1 Het2
r
asked Mar 28 at 20:57
LearnerLearner
4211 silver badge11 bronze badges
By "double points" do you mean a colon? Is this a regional term? Never heard it in the US
– camille
Mar 28 at 21:03
@camille In Portugal it is "dois pontos", meaning "two points".
– Rui Barradas
Mar 28 at 21:05
@Camille I meant :
– Learner
Mar 28 at 21:06
Will02and2matches to the same string?
– akrun
Mar 28 at 21:09
@akrun Yes 02 and 2 are the same
– Learner
Mar 28 at 21:09
add a comment
|
1
1
1
I have a data like this
df<- structure(list(df = structure(c(10L, 8L, 2L, 8L, 7L, 7L, 10L,
8L, 3L, 10L, 10L, 9L, 9L, 1L, 1L, 3L, 1L, 5L, 5L, 4L, 10L, 8L,
1L, 1L, 2L, 6L), .Label = c("-1:-1:2", "-1:2:-1", "-1:2:2", "1:01:01",
"1:1(2):1", "1(1)|1(2):1(1)|1(2):1(1)|1(2)", "1(1)|1(2):2:2",
"2:-1:-1", "2:-1:2", "2:02:02"), class = "factor")), class = "data.frame", row.names = c(NA,
-26L))
I want to expend it to words as I define. I want to have as many columns as the number of double points , for example here we have three : so we will add 3 columns after the df. Then we fill it up with words
2 = Homo
-1 = No
1= Het
1(1)= Het1
1(2)= Het2
So an expected output looks like below.
2:02:02 Homo Homo Homo
2:-1:-1 Homo No No
-1:2:-1 No Homo No
2:-1:-1 Homo No No
1(1)|1(2):2:2 Het1 Het2 Homo Homo
1(1)|1(2):2:2 Het1 Het2 Homo Homo
2:02:02 Homo Homo Homo
2:-1:-1 Homo No No
-1:2:2 No Homo Homo
2:02:02 Homo Homo Homo
2:02:02 Homo Homo Homo
2:-1:2 Homo No Homo
2:-1:2 Homo No Homo
-1:-1:2 No No Homo
-1:-1:2 No No Homo
-1:2:2 No Homo Homo
-1:-1:2 No No Homo
1:1(2):1 Het Het2 Het
1:1(2):1 Het Het3 Het
1:01:01 Het Het Het
2:02:02 Homo Homo Homo
2:-1:-1 Homo No No
-1:-1:2 No No Homo
-1:-1:2 No No Homo
-1:2:-1 No Homo No
1(1)|1(2):1(1)|1(2):1(1)|1(2) Het1 Het2 Het1 Het2 Het1 Het2
r
asked Mar 28 at 20:57
LearnerLearner
4211 silver badge11 bronze badges
I have a data like this
df<- structure(list(df = structure(c(10L, 8L, 2L, 8L, 7L, 7L, 10L,
8L, 3L, 10L, 10L, 9L, 9L, 1L, 1L, 3L, 1L, 5L, 5L, 4L, 10L, 8L,
1L, 1L, 2L, 6L), .Label = c("-1:-1:2", "-1:2:-1", "-1:2:2", "1:01:01",
"1:1(2):1", "1(1)|1(2):1(1)|1(2):1(1)|1(2)", "1(1)|1(2):2:2",
"2:-1:-1", "2:-1:2", "2:02:02"), class = "factor")), class = "data.frame", row.names = c(NA,
-26L))
I want to expend it to words as I define. I want to have as many columns as the number of double points , for example here we have three : so we will add 3 columns after the df. Then we fill it up with words
2 = Homo
-1 = No
1= Het
1(1)= Het1
1(2)= Het2
So an expected output looks like below.
2:02:02 Homo Homo Homo
2:-1:-1 Homo No No
-1:2:-1 No Homo No
2:-1:-1 Homo No No
1(1)|1(2):2:2 Het1 Het2 Homo Homo
1(1)|1(2):2:2 Het1 Het2 Homo Homo
2:02:02 Homo Homo Homo
2:-1:-1 Homo No No
-1:2:2 No Homo Homo
2:02:02 Homo Homo Homo
2:02:02 Homo Homo Homo
2:-1:2 Homo No Homo
2:-1:2 Homo No Homo
-1:-1:2 No No Homo
-1:-1:2 No No Homo
-1:2:2 No Homo Homo
-1:-1:2 No No Homo
1:1(2):1 Het Het2 Het
1:1(2):1 Het Het3 Het
1:01:01 Het Het Het
2:02:02 Homo Homo Homo
2:-1:-1 Homo No No
-1:-1:2 No No Homo
-1:-1:2 No No Homo
-1:2:-1 No Homo No
1(1)|1(2):1(1)|1(2):1(1)|1(2) Het1 Het2 Het1 Het2 Het1 Het2
r
r
asked Mar 28 at 20:57
LearnerLearner
4211 silver badge11 bronze badges
asked Mar 28 at 20:57
LearnerLearner
4211 silver badge11 bronze badges
edited Mar 28 at 21:08
Learner
asked Mar 28 at 20:57
LearnerLearner
4211 silver badge11 bronze badges
asked Mar 28 at 20:57
LearnerLearner
4211 silver badge11 bronze badges
asked Mar 28 at 20:57
LearnerLearner
4211 silver badge11 bronze badges
4211 silver badge11 bronze badges
By "double points" do you mean a colon? Is this a regional term? Never heard it in the US
– camille
Mar 28 at 21:03
@camille In Portugal it is "dois pontos", meaning "two points".
– Rui Barradas
Mar 28 at 21:05
@Camille I meant :
– Learner
Mar 28 at 21:06
Will02and2matches to the same string?
– akrun
Mar 28 at 21:09
@akrun Yes 02 and 2 are the same
– Learner
Mar 28 at 21:09
add a comment
|
By "double points" do you mean a colon? Is this a regional term? Never heard it in the US
– camille
Mar 28 at 21:03
@camille In Portugal it is "dois pontos", meaning "two points".
– Rui Barradas
Mar 28 at 21:05
@Camille I meant :
– Learner
Mar 28 at 21:06
Will02and2matches to the same string?
– akrun
Mar 28 at 21:09
@akrun Yes 02 and 2 are the same
– Learner
Mar 28 at 21:09
By "double points" do you mean a colon? Is this a regional term? Never heard it in the US
– camille
Mar 28 at 21:03
By "double points" do you mean a colon? Is this a regional term? Never heard it in the US
– camille
Mar 28 at 21:03
@camille In Portugal it is "dois pontos", meaning "two points".
– Rui Barradas
Mar 28 at 21:05
@camille In Portugal it is "dois pontos", meaning "two points".
– Rui Barradas
Mar 28 at 21:05
@Camille I meant :
– Learner
Mar 28 at 21:06
@Camille I meant :
– Learner
Mar 28 at 21:06
Will
02 and 2 matches to the same string?– akrun
Mar 28 at 21:09
Will
02 and 2 matches to the same string?– akrun
Mar 28 at 21:09
@akrun Yes 02 and 2 are the same
– Learner
Mar 28 at 21:09
@akrun Yes 02 and 2 are the same
– Learner
Mar 28 at 21:09
add a comment
|
2 Answers
2
active
oldest
votes
1
Not sure if the result is exactly what you need, but maybe this could help.
I think also maybe it's not the most efficient and beautiful solution, but it can be a starting point.
However, I called dats your data:
head(dats)
df
1 2:02:02
2 2:-1:-1
3 -1:2:-1
4 2:-1:-1
5 1(1)|1(2):2:2
6 1(1)|1(2):2:2
And I created a mapping data.frame:
mapping
id value
1 2 Homo
2 -1 No
3 1 Het
4 1(1) Het1
5 1(2) Het2
First, I splitted with stringr::str_split_fixed() the double points:
library(stringr)
double_point <- as.data.frame.matrix(str_split_fixed(dats$df, ":", 3))
Now we have to separate for each column, the values by |:
listed <- list() # empty list
for (i in (1:ncol(double_point)))
listed[[i]] <- (double_point[,i])
listed[[i]] <- str_split_fixed(listed[[i]], "\
# put as data frame
df_ <- do.call(cbind, listed)
# this is going to help in the future
df_1 <- df_
# result till now:
head(df_1)
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] "2" "" "02" "" "02" ""
[2,] "2" "" "-1" "" "-1" ""
[3,] "-1" "" "2" "" "-1" ""
[4,] "2" "" "-1" "" "-1" ""
[5,] "1(1)" "1(2)" "2" "" "2" ""
[6,] "1(1)" "1(2)" "2" "" "2" ""
Now we have to replace the values with the mapping, and bind them with the original data splitted (in this case):
listed <- list()
for (i in (1:ncol(df_)))
df_[,i] <- gsub("0","",df_[,i])
listed[[i]] <- mapping[match(df_[,i], mapping$id), 2, drop=F]
df_final <- cbind(df_1,do.call(cbind, listed))
head(df_final)
1 2 3 4 5 6 value value value value value value
1 2 02 02 Homo <NA> Homo <NA> Homo <NA>
1.1 2 -1 -1 Homo <NA> No <NA> No <NA>
2 -1 2 -1 No <NA> Homo <NA> No <NA>
1.2 2 -1 -1 Homo <NA> No <NA> No <NA>
4 1(1) 1(2) 2 2 Het1 Het2 Homo <NA> Homo <NA>
4.1 1(1) 1(2) 2 2 Het1 Het2 Homo <NA> Homo <NA>
Hope it helps!
EDIT
Here the mapping dput() and str():
dput(mapping)
structure(list(id = structure(c(5L, 1L, 2L, 3L, 4L), .Label = c("-1",
"1", "1(1)", "1(2)", "2"), class = "factor"), value = structure(c(4L,
5L, 1L, 2L, 3L), .Label = c("Het", "Het1", "Het2", "Homo", "No"
), class = "factor")), class = "data.frame", row.names = c("1",
"2", "3", "4", "5"))
str(mapping)
'data.frame': 5 obs. of 2 variables:
$ id : Factor w/ 5 levels "-1","1","1(1)",..: 5 1 2 3 4
$ value: Factor w/ 5 levels "Het","Het1","Het2",..: 4 5 1 2 3
answered Mar 29 at 10:32
s_ts_t
4,7742 gold badges12 silver badges35 bronze badges
Thanks, your code does not print for 1(1) or 1(2) . can you please tell me the str of mapping ?
– Learner
Mar 29 at 14:38
Hi,posted the edit. It seems that in the last rows of the last output, it prints for the cases you mention.
– s_t
Mar 29 at 15:10
1
I accepted and liked your answer
– Learner
Mar 29 at 20:02
add a comment
|
1
You can explicitly define all the possible values in num2words data frame and then run the following
df<- structure(list(df = structure(c(10L, 8L, 2L, 8L, 7L, 7L, 10L,
8L, 3L, 10L, 10L, 9L, 9L, 1L, 1L, 3L, 1L, 5L, 5L, 4L, 10L, 8L,
1L, 1L, 2L, 6L), .Label = c("-1:-1:2", "-1:2:-1", "-1:2:2", "1:01:01",
"1:1(2):1", "1(1)|1(2):1(1)|1(2):1(1)|1(2)", "1(1)|1(2):2:2",
"2:-1:-1", "2:-1:2", "2:02:02"), class = "factor")), class = "data.frame", row.names = c(NA,
-26L))
num2words <- read.table(text = "
num word
2 Homo
02 Homo
-1 No
1 Het
01 Het
1(1) Het1
1(2) Het2
1(1)|1(2) Het1-Het2
1(2)|1(1) Het2-Het1
", header = T, stringsAsFactors = F)
lst=lapply(1:nrow(df), function(x)
split.nums <- unlist(strsplit(as.character(df[x,]), ":"))
num2words$word[match(split.nums, num2words$num)]
)
new.df=cbind(df, do.call(rbind, lst))
> new.df
df 1 2 3
1 2:02:02 Homo Homo Homo
2 2:-1:-1 Homo No No
3 -1:2:-1 No Homo No
4 2:-1:-1 Homo No No
5 1(1)|1(2):2:2 Het1-Het2 Homo Homo
6 1(1)|1(2):2:2 Het1-Het2 Homo Homo
7 2:02:02 Homo Homo Homo
8 2:-1:-1 Homo No No
9 -1:2:2 No Homo Homo
10 2:02:02 Homo Homo Homo
11 2:02:02 Homo Homo Homo
12 2:-1:2 Homo No Homo
13 2:-1:2 Homo No Homo
14 -1:-1:2 No No Homo
15 -1:-1:2 No No Homo
16 -1:2:2 No Homo Homo
17 -1:-1:2 No No Homo
18 1:1(2):1 Het Het2 Het
19 1:1(2):1 Het Het2 Het
20 1:01:01 Het Het Het
21 2:02:02 Homo Homo Homo
22 2:-1:-1 Homo No No
23 -1:-1:2 No No Homo
24 -1:-1:2 No No Homo
25 -1:2:-1 No Homo No
26 1(1)|1(2):1(1)|1(2):1(1)|1(2) Het1-Het2 Het1-Het2 Het1-Het2
answered Mar 29 at 16:39
SinghTheCoderSinghTheCoder
1,38710 silver badges21 bronze badges
I liked your answer thank you
– Learner
Mar 29 at 20:02
Thanks. I think it is succinct and easy to read.
– SinghTheCoder
Mar 29 at 23:45
add a comment
|
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
Not sure if the result is exactly what you need, but maybe this could help.
I think also maybe it's not the most efficient and beautiful solution, but it can be a starting point.
However, I called dats your data:
head(dats)
df
1 2:02:02
2 2:-1:-1
3 -1:2:-1
4 2:-1:-1
5 1(1)|1(2):2:2
6 1(1)|1(2):2:2
And I created a mapping data.frame:
mapping
id value
1 2 Homo
2 -1 No
3 1 Het
4 1(1) Het1
5 1(2) Het2
First, I splitted with stringr::str_split_fixed() the double points:
library(stringr)
double_point <- as.data.frame.matrix(str_split_fixed(dats$df, ":", 3))
Now we have to separate for each column, the values by |:
listed <- list() # empty list
for (i in (1:ncol(double_point)))
listed[[i]] <- (double_point[,i])
listed[[i]] <- str_split_fixed(listed[[i]], "\
# put as data frame
df_ <- do.call(cbind, listed)
# this is going to help in the future
df_1 <- df_
# result till now:
head(df_1)
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] "2" "" "02" "" "02" ""
[2,] "2" "" "-1" "" "-1" ""
[3,] "-1" "" "2" "" "-1" ""
[4,] "2" "" "-1" "" "-1" ""
[5,] "1(1)" "1(2)" "2" "" "2" ""
[6,] "1(1)" "1(2)" "2" "" "2" ""
Now we have to replace the values with the mapping, and bind them with the original data splitted (in this case):
listed <- list()
for (i in (1:ncol(df_)))
df_[,i] <- gsub("0","",df_[,i])
listed[[i]] <- mapping[match(df_[,i], mapping$id), 2, drop=F]
df_final <- cbind(df_1,do.call(cbind, listed))
head(df_final)
1 2 3 4 5 6 value value value value value value
1 2 02 02 Homo <NA> Homo <NA> Homo <NA>
1.1 2 -1 -1 Homo <NA> No <NA> No <NA>
2 -1 2 -1 No <NA> Homo <NA> No <NA>
1.2 2 -1 -1 Homo <NA> No <NA> No <NA>
4 1(1) 1(2) 2 2 Het1 Het2 Homo <NA> Homo <NA>
4.1 1(1) 1(2) 2 2 Het1 Het2 Homo <NA> Homo <NA>
Hope it helps!
EDIT
Here the mapping dput() and str():
dput(mapping)
structure(list(id = structure(c(5L, 1L, 2L, 3L, 4L), .Label = c("-1",
"1", "1(1)", "1(2)", "2"), class = "factor"), value = structure(c(4L,
5L, 1L, 2L, 3L), .Label = c("Het", "Het1", "Het2", "Homo", "No"
), class = "factor")), class = "data.frame", row.names = c("1",
"2", "3", "4", "5"))
str(mapping)
'data.frame': 5 obs. of 2 variables:
$ id : Factor w/ 5 levels "-1","1","1(1)",..: 5 1 2 3 4
$ value: Factor w/ 5 levels "Het","Het1","Het2",..: 4 5 1 2 3
answered Mar 29 at 10:32
s_ts_t
4,7742 gold badges12 silver badges35 bronze badges
Thanks, your code does not print for 1(1) or 1(2) . can you please tell me the str of mapping ?
– Learner
Mar 29 at 14:38
Hi,posted the edit. It seems that in the last rows of the last output, it prints for the cases you mention.
– s_t
Mar 29 at 15:10
1
I accepted and liked your answer
– Learner
Mar 29 at 20:02
add a comment
|
1
Not sure if the result is exactly what you need, but maybe this could help.
I think also maybe it's not the most efficient and beautiful solution, but it can be a starting point.
However, I called dats your data:
head(dats)
df
1 2:02:02
2 2:-1:-1
3 -1:2:-1
4 2:-1:-1
5 1(1)|1(2):2:2
6 1(1)|1(2):2:2
And I created a mapping data.frame:
mapping
id value
1 2 Homo
2 -1 No
3 1 Het
4 1(1) Het1
5 1(2) Het2
First, I splitted with stringr::str_split_fixed() the double points:
library(stringr)
double_point <- as.data.frame.matrix(str_split_fixed(dats$df, ":", 3))
Now we have to separate for each column, the values by |:
listed <- list() # empty list
for (i in (1:ncol(double_point)))
listed[[i]] <- (double_point[,i])
listed[[i]] <- str_split_fixed(listed[[i]], "\
# put as data frame
df_ <- do.call(cbind, listed)
# this is going to help in the future
df_1 <- df_
# result till now:
head(df_1)
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] "2" "" "02" "" "02" ""
[2,] "2" "" "-1" "" "-1" ""
[3,] "-1" "" "2" "" "-1" ""
[4,] "2" "" "-1" "" "-1" ""
[5,] "1(1)" "1(2)" "2" "" "2" ""
[6,] "1(1)" "1(2)" "2" "" "2" ""
Now we have to replace the values with the mapping, and bind them with the original data splitted (in this case):
listed <- list()
for (i in (1:ncol(df_)))
df_[,i] <- gsub("0","",df_[,i])
listed[[i]] <- mapping[match(df_[,i], mapping$id), 2, drop=F]
df_final <- cbind(df_1,do.call(cbind, listed))
head(df_final)
1 2 3 4 5 6 value value value value value value
1 2 02 02 Homo <NA> Homo <NA> Homo <NA>
1.1 2 -1 -1 Homo <NA> No <NA> No <NA>
2 -1 2 -1 No <NA> Homo <NA> No <NA>
1.2 2 -1 -1 Homo <NA> No <NA> No <NA>
4 1(1) 1(2) 2 2 Het1 Het2 Homo <NA> Homo <NA>
4.1 1(1) 1(2) 2 2 Het1 Het2 Homo <NA> Homo <NA>
Hope it helps!
EDIT
Here the mapping dput() and str():
dput(mapping)
structure(list(id = structure(c(5L, 1L, 2L, 3L, 4L), .Label = c("-1",
"1", "1(1)", "1(2)", "2"), class = "factor"), value = structure(c(4L,
5L, 1L, 2L, 3L), .Label = c("Het", "Het1", "Het2", "Homo", "No"
), class = "factor")), class = "data.frame", row.names = c("1",
"2", "3", "4", "5"))
str(mapping)
'data.frame': 5 obs. of 2 variables:
$ id : Factor w/ 5 levels "-1","1","1(1)",..: 5 1 2 3 4
$ value: Factor w/ 5 levels "Het","Het1","Het2",..: 4 5 1 2 3
answered Mar 29 at 10:32
s_ts_t
4,7742 gold badges12 silver badges35 bronze badges
Thanks, your code does not print for 1(1) or 1(2) . can you please tell me the str of mapping ?
– Learner
Mar 29 at 14:38
Hi,posted the edit. It seems that in the last rows of the last output, it prints for the cases you mention.
– s_t
Mar 29 at 15:10
1
I accepted and liked your answer
– Learner
Mar 29 at 20:02
add a comment
|
1
1
1
Not sure if the result is exactly what you need, but maybe this could help.
I think also maybe it's not the most efficient and beautiful solution, but it can be a starting point.
However, I called dats your data:
head(dats)
df
1 2:02:02
2 2:-1:-1
3 -1:2:-1
4 2:-1:-1
5 1(1)|1(2):2:2
6 1(1)|1(2):2:2
And I created a mapping data.frame:
mapping
id value
1 2 Homo
2 -1 No
3 1 Het
4 1(1) Het1
5 1(2) Het2
First, I splitted with stringr::str_split_fixed() the double points:
library(stringr)
double_point <- as.data.frame.matrix(str_split_fixed(dats$df, ":", 3))
Now we have to separate for each column, the values by |:
listed <- list() # empty list
for (i in (1:ncol(double_point)))
listed[[i]] <- (double_point[,i])
listed[[i]] <- str_split_fixed(listed[[i]], "\
# put as data frame
df_ <- do.call(cbind, listed)
# this is going to help in the future
df_1 <- df_
# result till now:
head(df_1)
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] "2" "" "02" "" "02" ""
[2,] "2" "" "-1" "" "-1" ""
[3,] "-1" "" "2" "" "-1" ""
[4,] "2" "" "-1" "" "-1" ""
[5,] "1(1)" "1(2)" "2" "" "2" ""
[6,] "1(1)" "1(2)" "2" "" "2" ""
Now we have to replace the values with the mapping, and bind them with the original data splitted (in this case):
listed <- list()
for (i in (1:ncol(df_)))
df_[,i] <- gsub("0","",df_[,i])
listed[[i]] <- mapping[match(df_[,i], mapping$id), 2, drop=F]
df_final <- cbind(df_1,do.call(cbind, listed))
head(df_final)
1 2 3 4 5 6 value value value value value value
1 2 02 02 Homo <NA> Homo <NA> Homo <NA>
1.1 2 -1 -1 Homo <NA> No <NA> No <NA>
2 -1 2 -1 No <NA> Homo <NA> No <NA>
1.2 2 -1 -1 Homo <NA> No <NA> No <NA>
4 1(1) 1(2) 2 2 Het1 Het2 Homo <NA> Homo <NA>
4.1 1(1) 1(2) 2 2 Het1 Het2 Homo <NA> Homo <NA>
Hope it helps!
EDIT
Here the mapping dput() and str():
dput(mapping)
structure(list(id = structure(c(5L, 1L, 2L, 3L, 4L), .Label = c("-1",
"1", "1(1)", "1(2)", "2"), class = "factor"), value = structure(c(4L,
5L, 1L, 2L, 3L), .Label = c("Het", "Het1", "Het2", "Homo", "No"
), class = "factor")), class = "data.frame", row.names = c("1",
"2", "3", "4", "5"))
str(mapping)
'data.frame': 5 obs. of 2 variables:
$ id : Factor w/ 5 levels "-1","1","1(1)",..: 5 1 2 3 4
$ value: Factor w/ 5 levels "Het","Het1","Het2",..: 4 5 1 2 3
answered Mar 29 at 10:32
s_ts_t
4,7742 gold badges12 silver badges35 bronze badges
Not sure if the result is exactly what you need, but maybe this could help.
I think also maybe it's not the most efficient and beautiful solution, but it can be a starting point.
However, I called dats your data:
head(dats)
df
1 2:02:02
2 2:-1:-1
3 -1:2:-1
4 2:-1:-1
5 1(1)|1(2):2:2
6 1(1)|1(2):2:2
And I created a mapping data.frame:
mapping
id value
1 2 Homo
2 -1 No
3 1 Het
4 1(1) Het1
5 1(2) Het2
First, I splitted with stringr::str_split_fixed() the double points:
library(stringr)
double_point <- as.data.frame.matrix(str_split_fixed(dats$df, ":", 3))
Now we have to separate for each column, the values by |:
listed <- list() # empty list
for (i in (1:ncol(double_point)))
listed[[i]] <- (double_point[,i])
listed[[i]] <- str_split_fixed(listed[[i]], "\
# put as data frame
df_ <- do.call(cbind, listed)
# this is going to help in the future
df_1 <- df_
# result till now:
head(df_1)
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] "2" "" "02" "" "02" ""
[2,] "2" "" "-1" "" "-1" ""
[3,] "-1" "" "2" "" "-1" ""
[4,] "2" "" "-1" "" "-1" ""
[5,] "1(1)" "1(2)" "2" "" "2" ""
[6,] "1(1)" "1(2)" "2" "" "2" ""
Now we have to replace the values with the mapping, and bind them with the original data splitted (in this case):
listed <- list()
for (i in (1:ncol(df_)))
df_[,i] <- gsub("0","",df_[,i])
listed[[i]] <- mapping[match(df_[,i], mapping$id), 2, drop=F]
df_final <- cbind(df_1,do.call(cbind, listed))
head(df_final)
1 2 3 4 5 6 value value value value value value
1 2 02 02 Homo <NA> Homo <NA> Homo <NA>
1.1 2 -1 -1 Homo <NA> No <NA> No <NA>
2 -1 2 -1 No <NA> Homo <NA> No <NA>
1.2 2 -1 -1 Homo <NA> No <NA> No <NA>
4 1(1) 1(2) 2 2 Het1 Het2 Homo <NA> Homo <NA>
4.1 1(1) 1(2) 2 2 Het1 Het2 Homo <NA> Homo <NA>
Hope it helps!
EDIT
Here the mapping dput() and str():
dput(mapping)
structure(list(id = structure(c(5L, 1L, 2L, 3L, 4L), .Label = c("-1",
"1", "1(1)", "1(2)", "2"), class = "factor"), value = structure(c(4L,
5L, 1L, 2L, 3L), .Label = c("Het", "Het1", "Het2", "Homo", "No"
), class = "factor")), class = "data.frame", row.names = c("1",
"2", "3", "4", "5"))
str(mapping)
'data.frame': 5 obs. of 2 variables:
$ id : Factor w/ 5 levels "-1","1","1(1)",..: 5 1 2 3 4
$ value: Factor w/ 5 levels "Het","Het1","Het2",..: 4 5 1 2 3
answered Mar 29 at 10:32
s_ts_t
4,7742 gold badges12 silver badges35 bronze badges
edited Mar 29 at 15:09
answered Mar 29 at 10:32
s_ts_t
4,7742 gold badges12 silver badges35 bronze badges
answered Mar 29 at 10:32
s_ts_t
4,7742 gold badges12 silver badges35 bronze badges
answered Mar 29 at 10:32
s_ts_t
4,7742 gold badges12 silver badges35 bronze badges
4,7742 gold badges12 silver badges35 bronze badges
Thanks, your code does not print for 1(1) or 1(2) . can you please tell me the str of mapping ?
– Learner
Mar 29 at 14:38
Hi,posted the edit. It seems that in the last rows of the last output, it prints for the cases you mention.
– s_t
Mar 29 at 15:10
1
I accepted and liked your answer
– Learner
Mar 29 at 20:02
add a comment
|
Thanks, your code does not print for 1(1) or 1(2) . can you please tell me the str of mapping ?
– Learner
Mar 29 at 14:38
Hi,posted the edit. It seems that in the last rows of the last output, it prints for the cases you mention.
– s_t
Mar 29 at 15:10
1
I accepted and liked your answer
– Learner
Mar 29 at 20:02
Thanks, your code does not print for 1(1) or 1(2) . can you please tell me the str of mapping ?
– Learner
Mar 29 at 14:38
Thanks, your code does not print for 1(1) or 1(2) . can you please tell me the str of mapping ?
– Learner
Mar 29 at 14:38
Hi,posted the edit. It seems that in the last rows of the last output, it prints for the cases you mention.
– s_t
Mar 29 at 15:10
Hi,posted the edit. It seems that in the last rows of the last output, it prints for the cases you mention.
– s_t
Mar 29 at 15:10
1
1
I accepted and liked your answer
– Learner
Mar 29 at 20:02
I accepted and liked your answer
– Learner
Mar 29 at 20:02
add a comment
|
1
You can explicitly define all the possible values in num2words data frame and then run the following
df<- structure(list(df = structure(c(10L, 8L, 2L, 8L, 7L, 7L, 10L,
8L, 3L, 10L, 10L, 9L, 9L, 1L, 1L, 3L, 1L, 5L, 5L, 4L, 10L, 8L,
1L, 1L, 2L, 6L), .Label = c("-1:-1:2", "-1:2:-1", "-1:2:2", "1:01:01",
"1:1(2):1", "1(1)|1(2):1(1)|1(2):1(1)|1(2)", "1(1)|1(2):2:2",
"2:-1:-1", "2:-1:2", "2:02:02"), class = "factor")), class = "data.frame", row.names = c(NA,
-26L))
num2words <- read.table(text = "
num word
2 Homo
02 Homo
-1 No
1 Het
01 Het
1(1) Het1
1(2) Het2
1(1)|1(2) Het1-Het2
1(2)|1(1) Het2-Het1
", header = T, stringsAsFactors = F)
lst=lapply(1:nrow(df), function(x)
split.nums <- unlist(strsplit(as.character(df[x,]), ":"))
num2words$word[match(split.nums, num2words$num)]
)
new.df=cbind(df, do.call(rbind, lst))
> new.df
df 1 2 3
1 2:02:02 Homo Homo Homo
2 2:-1:-1 Homo No No
3 -1:2:-1 No Homo No
4 2:-1:-1 Homo No No
5 1(1)|1(2):2:2 Het1-Het2 Homo Homo
6 1(1)|1(2):2:2 Het1-Het2 Homo Homo
7 2:02:02 Homo Homo Homo
8 2:-1:-1 Homo No No
9 -1:2:2 No Homo Homo
10 2:02:02 Homo Homo Homo
11 2:02:02 Homo Homo Homo
12 2:-1:2 Homo No Homo
13 2:-1:2 Homo No Homo
14 -1:-1:2 No No Homo
15 -1:-1:2 No No Homo
16 -1:2:2 No Homo Homo
17 -1:-1:2 No No Homo
18 1:1(2):1 Het Het2 Het
19 1:1(2):1 Het Het2 Het
20 1:01:01 Het Het Het
21 2:02:02 Homo Homo Homo
22 2:-1:-1 Homo No No
23 -1:-1:2 No No Homo
24 -1:-1:2 No No Homo
25 -1:2:-1 No Homo No
26 1(1)|1(2):1(1)|1(2):1(1)|1(2) Het1-Het2 Het1-Het2 Het1-Het2
answered Mar 29 at 16:39
SinghTheCoderSinghTheCoder
1,38710 silver badges21 bronze badges
I liked your answer thank you
– Learner
Mar 29 at 20:02
Thanks. I think it is succinct and easy to read.
– SinghTheCoder
Mar 29 at 23:45
add a comment
|
1
You can explicitly define all the possible values in num2words data frame and then run the following
df<- structure(list(df = structure(c(10L, 8L, 2L, 8L, 7L, 7L, 10L,
8L, 3L, 10L, 10L, 9L, 9L, 1L, 1L, 3L, 1L, 5L, 5L, 4L, 10L, 8L,
1L, 1L, 2L, 6L), .Label = c("-1:-1:2", "-1:2:-1", "-1:2:2", "1:01:01",
"1:1(2):1", "1(1)|1(2):1(1)|1(2):1(1)|1(2)", "1(1)|1(2):2:2",
"2:-1:-1", "2:-1:2", "2:02:02"), class = "factor")), class = "data.frame", row.names = c(NA,
-26L))
num2words <- read.table(text = "
num word
2 Homo
02 Homo
-1 No
1 Het
01 Het
1(1) Het1
1(2) Het2
1(1)|1(2) Het1-Het2
1(2)|1(1) Het2-Het1
", header = T, stringsAsFactors = F)
lst=lapply(1:nrow(df), function(x)
split.nums <- unlist(strsplit(as.character(df[x,]), ":"))
num2words$word[match(split.nums, num2words$num)]
)
new.df=cbind(df, do.call(rbind, lst))
> new.df
df 1 2 3
1 2:02:02 Homo Homo Homo
2 2:-1:-1 Homo No No
3 -1:2:-1 No Homo No
4 2:-1:-1 Homo No No
5 1(1)|1(2):2:2 Het1-Het2 Homo Homo
6 1(1)|1(2):2:2 Het1-Het2 Homo Homo
7 2:02:02 Homo Homo Homo
8 2:-1:-1 Homo No No
9 -1:2:2 No Homo Homo
10 2:02:02 Homo Homo Homo
11 2:02:02 Homo Homo Homo
12 2:-1:2 Homo No Homo
13 2:-1:2 Homo No Homo
14 -1:-1:2 No No Homo
15 -1:-1:2 No No Homo
16 -1:2:2 No Homo Homo
17 -1:-1:2 No No Homo
18 1:1(2):1 Het Het2 Het
19 1:1(2):1 Het Het2 Het
20 1:01:01 Het Het Het
21 2:02:02 Homo Homo Homo
22 2:-1:-1 Homo No No
23 -1:-1:2 No No Homo
24 -1:-1:2 No No Homo
25 -1:2:-1 No Homo No
26 1(1)|1(2):1(1)|1(2):1(1)|1(2) Het1-Het2 Het1-Het2 Het1-Het2
answered Mar 29 at 16:39
SinghTheCoderSinghTheCoder
1,38710 silver badges21 bronze badges
I liked your answer thank you
– Learner
Mar 29 at 20:02
Thanks. I think it is succinct and easy to read.
– SinghTheCoder
Mar 29 at 23:45
add a comment
|
1
1
1
You can explicitly define all the possible values in num2words data frame and then run the following
df<- structure(list(df = structure(c(10L, 8L, 2L, 8L, 7L, 7L, 10L,
8L, 3L, 10L, 10L, 9L, 9L, 1L, 1L, 3L, 1L, 5L, 5L, 4L, 10L, 8L,
1L, 1L, 2L, 6L), .Label = c("-1:-1:2", "-1:2:-1", "-1:2:2", "1:01:01",
"1:1(2):1", "1(1)|1(2):1(1)|1(2):1(1)|1(2)", "1(1)|1(2):2:2",
"2:-1:-1", "2:-1:2", "2:02:02"), class = "factor")), class = "data.frame", row.names = c(NA,
-26L))
num2words <- read.table(text = "
num word
2 Homo
02 Homo
-1 No
1 Het
01 Het
1(1) Het1
1(2) Het2
1(1)|1(2) Het1-Het2
1(2)|1(1) Het2-Het1
", header = T, stringsAsFactors = F)
lst=lapply(1:nrow(df), function(x)
split.nums <- unlist(strsplit(as.character(df[x,]), ":"))
num2words$word[match(split.nums, num2words$num)]
)
new.df=cbind(df, do.call(rbind, lst))
> new.df
df 1 2 3
1 2:02:02 Homo Homo Homo
2 2:-1:-1 Homo No No
3 -1:2:-1 No Homo No
4 2:-1:-1 Homo No No
5 1(1)|1(2):2:2 Het1-Het2 Homo Homo
6 1(1)|1(2):2:2 Het1-Het2 Homo Homo
7 2:02:02 Homo Homo Homo
8 2:-1:-1 Homo No No
9 -1:2:2 No Homo Homo
10 2:02:02 Homo Homo Homo
11 2:02:02 Homo Homo Homo
12 2:-1:2 Homo No Homo
13 2:-1:2 Homo No Homo
14 -1:-1:2 No No Homo
15 -1:-1:2 No No Homo
16 -1:2:2 No Homo Homo
17 -1:-1:2 No No Homo
18 1:1(2):1 Het Het2 Het
19 1:1(2):1 Het Het2 Het
20 1:01:01 Het Het Het
21 2:02:02 Homo Homo Homo
22 2:-1:-1 Homo No No
23 -1:-1:2 No No Homo
24 -1:-1:2 No No Homo
25 -1:2:-1 No Homo No
26 1(1)|1(2):1(1)|1(2):1(1)|1(2) Het1-Het2 Het1-Het2 Het1-Het2
answered Mar 29 at 16:39
SinghTheCoderSinghTheCoder
1,38710 silver badges21 bronze badges
You can explicitly define all the possible values in num2words data frame and then run the following
df<- structure(list(df = structure(c(10L, 8L, 2L, 8L, 7L, 7L, 10L,
8L, 3L, 10L, 10L, 9L, 9L, 1L, 1L, 3L, 1L, 5L, 5L, 4L, 10L, 8L,
1L, 1L, 2L, 6L), .Label = c("-1:-1:2", "-1:2:-1", "-1:2:2", "1:01:01",
"1:1(2):1", "1(1)|1(2):1(1)|1(2):1(1)|1(2)", "1(1)|1(2):2:2",
"2:-1:-1", "2:-1:2", "2:02:02"), class = "factor")), class = "data.frame", row.names = c(NA,
-26L))
num2words <- read.table(text = "
num word
2 Homo
02 Homo
-1 No
1 Het
01 Het
1(1) Het1
1(2) Het2
1(1)|1(2) Het1-Het2
1(2)|1(1) Het2-Het1
", header = T, stringsAsFactors = F)
lst=lapply(1:nrow(df), function(x)
split.nums <- unlist(strsplit(as.character(df[x,]), ":"))
num2words$word[match(split.nums, num2words$num)]
)
new.df=cbind(df, do.call(rbind, lst))
> new.df
df 1 2 3
1 2:02:02 Homo Homo Homo
2 2:-1:-1 Homo No No
3 -1:2:-1 No Homo No
4 2:-1:-1 Homo No No
5 1(1)|1(2):2:2 Het1-Het2 Homo Homo
6 1(1)|1(2):2:2 Het1-Het2 Homo Homo
7 2:02:02 Homo Homo Homo
8 2:-1:-1 Homo No No
9 -1:2:2 No Homo Homo
10 2:02:02 Homo Homo Homo
11 2:02:02 Homo Homo Homo
12 2:-1:2 Homo No Homo
13 2:-1:2 Homo No Homo
14 -1:-1:2 No No Homo
15 -1:-1:2 No No Homo
16 -1:2:2 No Homo Homo
17 -1:-1:2 No No Homo
18 1:1(2):1 Het Het2 Het
19 1:1(2):1 Het Het2 Het
20 1:01:01 Het Het Het
21 2:02:02 Homo Homo Homo
22 2:-1:-1 Homo No No
23 -1:-1:2 No No Homo
24 -1:-1:2 No No Homo
25 -1:2:-1 No Homo No
26 1(1)|1(2):1(1)|1(2):1(1)|1(2) Het1-Het2 Het1-Het2 Het1-Het2
answered Mar 29 at 16:39
SinghTheCoderSinghTheCoder
1,38710 silver badges21 bronze badges
answered Mar 29 at 16:39
SinghTheCoderSinghTheCoder
1,38710 silver badges21 bronze badges
answered Mar 29 at 16:39
SinghTheCoderSinghTheCoder
1,38710 silver badges21 bronze badges
answered Mar 29 at 16:39
SinghTheCoderSinghTheCoder
1,38710 silver badges21 bronze badges
1,38710 silver badges21 bronze badges
I liked your answer thank you
– Learner
Mar 29 at 20:02
Thanks. I think it is succinct and easy to read.
– SinghTheCoder
Mar 29 at 23:45
add a comment
|
I liked your answer thank you
– Learner
Mar 29 at 20:02
Thanks. I think it is succinct and easy to read.
– SinghTheCoder
Mar 29 at 23:45
I liked your answer thank you
– Learner
Mar 29 at 20:02
I liked your answer thank you
– Learner
Mar 29 at 20:02
Thanks. I think it is succinct and easy to read.
– SinghTheCoder
Mar 29 at 23:45
Thanks. I think it is succinct and easy to read.
– SinghTheCoder
Mar 29 at 23:45
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By "double points" do you mean a colon? Is this a regional term? Never heard it in the US
– camille
Mar 28 at 21:03
@camille In Portugal it is "dois pontos", meaning "two points".
– Rui Barradas
Mar 28 at 21:05
@Camille I meant :
– Learner
Mar 28 at 21:06
Will
02and2matches to the same string?– akrun
Mar 28 at 21:09
@akrun Yes 02 and 2 are the same
– Learner
Mar 28 at 21:09