How to extract lower and upper bound in numeric format from a confidence interval string?How can I plot data with confidence intervals?R - Bootstrapped Confidence Interval - Obtain Parameters of Upper and Lower BoundsHow to display several bootstrap confidence intervals on one plotHow can confidence intervals be numerically visualized in corrplot()?How to compute confidence Interval of the fitted value via nls()rAmCharts Scattered plot with Upper and lower confidence interval around Fitted valuesextract confidence intervals from binomial test in Rhow to add confidence interval as coloured area to a boxplot of predicted values in ggplot2?How to extract values of confidence interval from ggplot2 stat summary in R?Plotting a ggplot() when having the confidence interval upper and lower values
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How to extract lower and upper bound in numeric format from a confidence interval string?
How can I plot data with confidence intervals?R - Bootstrapped Confidence Interval - Obtain Parameters of Upper and Lower BoundsHow to display several bootstrap confidence intervals on one plotHow can confidence intervals be numerically visualized in corrplot()?How to compute confidence Interval of the fitted value via nls()rAmCharts Scattered plot with Upper and lower confidence interval around Fitted valuesextract confidence intervals from binomial test in Rhow to add confidence interval as coloured area to a boxplot of predicted values in ggplot2?How to extract values of confidence interval from ggplot2 stat summary in R?Plotting a ggplot() when having the confidence interval upper and lower values
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty
margin-bottom:0;
Suppose a vector including some confidence intervals as below
confint <- c("[0.741 ; 2.233]", "[263.917 ; 402.154]", "[12.788 ; 17.975]", "[0.680 ; 2.450]", "[0.650 ; 1.827]", "[0.719 ; 2.190]")
I want to have two new vectors one including the lower Limits in numeric format as
lower <- c(0.741, 263.917, 12.788, 0.680, 0.650 , 0.719)
and othe including the upper Limits in numeric format like
upper <- c(2.233, 402.154, 17.975, 2.450, 1.827, 2.190)
r
asked May 27 at 12:31
FatetaFateta
1298 bronze badges
add a comment
|
Suppose a vector including some confidence intervals as below
confint <- c("[0.741 ; 2.233]", "[263.917 ; 402.154]", "[12.788 ; 17.975]", "[0.680 ; 2.450]", "[0.650 ; 1.827]", "[0.719 ; 2.190]")
I want to have two new vectors one including the lower Limits in numeric format as
lower <- c(0.741, 263.917, 12.788, 0.680, 0.650 , 0.719)
and othe including the upper Limits in numeric format like
upper <- c(2.233, 402.154, 17.975, 2.450, 1.827, 2.190)
r
asked May 27 at 12:31
FatetaFateta
1298 bronze badges
add a comment
|
Suppose a vector including some confidence intervals as below
confint <- c("[0.741 ; 2.233]", "[263.917 ; 402.154]", "[12.788 ; 17.975]", "[0.680 ; 2.450]", "[0.650 ; 1.827]", "[0.719 ; 2.190]")
I want to have two new vectors one including the lower Limits in numeric format as
lower <- c(0.741, 263.917, 12.788, 0.680, 0.650 , 0.719)
and othe including the upper Limits in numeric format like
upper <- c(2.233, 402.154, 17.975, 2.450, 1.827, 2.190)
r
asked May 27 at 12:31
FatetaFateta
1298 bronze badges
Suppose a vector including some confidence intervals as below
confint <- c("[0.741 ; 2.233]", "[263.917 ; 402.154]", "[12.788 ; 17.975]", "[0.680 ; 2.450]", "[0.650 ; 1.827]", "[0.719 ; 2.190]")
I want to have two new vectors one including the lower Limits in numeric format as
lower <- c(0.741, 263.917, 12.788, 0.680, 0.650 , 0.719)
and othe including the upper Limits in numeric format like
upper <- c(2.233, 402.154, 17.975, 2.450, 1.827, 2.190)
r
r
asked May 27 at 12:31
FatetaFateta
1298 bronze badges
asked May 27 at 12:31
FatetaFateta
1298 bronze badges
asked May 27 at 12:31
FatetaFateta
1298 bronze badges
asked May 27 at 12:31
FatetaFateta
1298 bronze badges
asked May 27 at 12:31
FatetaFateta
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6 Answers
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A base R solution
lower = as.numeric(sub(".*?(\d+\.\d+).*", "\1", confint))
upper = as.numeric(sub(".*\b(\d+\.\d+).*", "\1", confint))
lower
[1] 0.741 263.917 12.788 0.680 0.650 0.719
upper
[1] 2.233 402.154 17.975 2.450 1.827 2.190
answered May 27 at 12:41
G5WG5W
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mypattern <- '\[(\d+\.\d+) ; (\d+\.\d+)\]'
as.numeric(gsub(mypattern, '\1', confint))
as.numeric(gsub(mypattern, '\2', confint))
answered May 27 at 12:52
pzhaopzhao
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A different base R possibility could be:
sapply(strsplit(confint, " ; ", fixed = TRUE), function(x) gsub("[^0-9.-]+", "\1", x) [1])
sapply(strsplit(confint, " ; ", fixed = TRUE), function(x) gsub("[^0-9.-]+", "\1", x) [2])
[1] "0.741" "263.917" "12.788" "0.680" "0.650" "0.719"
[1] "2.233" "402.154" "17.975" "2.450" "1.827" "2.190"
If you need it as a numeric vector:
sapply(strsplit(confint, " ; ", fixed = TRUE), function(x) as.numeric(gsub("[^0-9.-]+", "\1", x)) [1])
sapply(strsplit(confint, " ; ", fixed = TRUE), function(x) as.numeric(gsub("[^0-9.-]+", "\1", x)) [2])
answered May 27 at 12:49
tmfmnktmfmnk
12.4k1 gold badge10 silver badges26 bronze badges
@Ronak Shah I updated my post according some of your ideas. Thank you :)
– tmfmnk
May 27 at 12:59
I realised it needed some more tweaking to get exact output but I think it was a similar approach. +1 to you.
– Ronak Shah
May 27 at 13:21
add a comment
|
Meanwhile, I came up with another base R solution
lower <- as.numeric(sub(".]*", "", sub(";.*", "", confint)))
upper <- as.numeric(sub("].*", "", sub(".*;", "", confint)))
Thank you all!
answered May 27 at 13:22
FatetaFateta
1298 bronze badges
add a comment
|
You can use functions from the stringr library.
You can split strings with str_split() according a specific character (; in your case), then remove character with str_remove() ([ and ] in your case) and you will obtain what you want.
str_remove(str_split_fixed(confint, ";", n = 2)[,1], '\[') %>% as.numeric()
# [1] 0.741 263.917 12.788 0.680 0.650 0.719
answered May 27 at 12:38
demarsylvaindemarsylvain
1,5132 gold badges7 silver badges21 bronze badges
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A tidyverse solution:
library(dplyr)
library(tidyr)
df = data.frame(confint)
df = df %>%
mutate(confint = gsub("(\[|\])","",confint)) %>%
separate(confint,c("lower","upper"),";",convert=T)
answered May 27 at 14:12
FinoFino
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6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
A base R solution
lower = as.numeric(sub(".*?(\d+\.\d+).*", "\1", confint))
upper = as.numeric(sub(".*\b(\d+\.\d+).*", "\1", confint))
lower
[1] 0.741 263.917 12.788 0.680 0.650 0.719
upper
[1] 2.233 402.154 17.975 2.450 1.827 2.190
answered May 27 at 12:41
G5WG5W
25.9k9 gold badges24 silver badges46 bronze badges
add a comment
|
A base R solution
lower = as.numeric(sub(".*?(\d+\.\d+).*", "\1", confint))
upper = as.numeric(sub(".*\b(\d+\.\d+).*", "\1", confint))
lower
[1] 0.741 263.917 12.788 0.680 0.650 0.719
upper
[1] 2.233 402.154 17.975 2.450 1.827 2.190
answered May 27 at 12:41
G5WG5W
25.9k9 gold badges24 silver badges46 bronze badges
add a comment
|
A base R solution
lower = as.numeric(sub(".*?(\d+\.\d+).*", "\1", confint))
upper = as.numeric(sub(".*\b(\d+\.\d+).*", "\1", confint))
lower
[1] 0.741 263.917 12.788 0.680 0.650 0.719
upper
[1] 2.233 402.154 17.975 2.450 1.827 2.190
answered May 27 at 12:41
G5WG5W
25.9k9 gold badges24 silver badges46 bronze badges
A base R solution
lower = as.numeric(sub(".*?(\d+\.\d+).*", "\1", confint))
upper = as.numeric(sub(".*\b(\d+\.\d+).*", "\1", confint))
lower
[1] 0.741 263.917 12.788 0.680 0.650 0.719
upper
[1] 2.233 402.154 17.975 2.450 1.827 2.190
answered May 27 at 12:41
G5WG5W
25.9k9 gold badges24 silver badges46 bronze badges
edited May 27 at 12:45
answered May 27 at 12:41
G5WG5W
25.9k9 gold badges24 silver badges46 bronze badges
answered May 27 at 12:41
G5WG5W
25.9k9 gold badges24 silver badges46 bronze badges
answered May 27 at 12:41
G5WG5W
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25.9k9 gold badges24 silver badges46 bronze badges
add a comment
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add a comment
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mypattern <- '\[(\d+\.\d+) ; (\d+\.\d+)\]'
as.numeric(gsub(mypattern, '\1', confint))
as.numeric(gsub(mypattern, '\2', confint))
answered May 27 at 12:52
pzhaopzhao
2401 silver badge8 bronze badges
add a comment
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mypattern <- '\[(\d+\.\d+) ; (\d+\.\d+)\]'
as.numeric(gsub(mypattern, '\1', confint))
as.numeric(gsub(mypattern, '\2', confint))
answered May 27 at 12:52
pzhaopzhao
2401 silver badge8 bronze badges
add a comment
|
mypattern <- '\[(\d+\.\d+) ; (\d+\.\d+)\]'
as.numeric(gsub(mypattern, '\1', confint))
as.numeric(gsub(mypattern, '\2', confint))
answered May 27 at 12:52
pzhaopzhao
2401 silver badge8 bronze badges
mypattern <- '\[(\d+\.\d+) ; (\d+\.\d+)\]'
as.numeric(gsub(mypattern, '\1', confint))
as.numeric(gsub(mypattern, '\2', confint))
answered May 27 at 12:52
pzhaopzhao
2401 silver badge8 bronze badges
answered May 27 at 12:52
pzhaopzhao
2401 silver badge8 bronze badges
answered May 27 at 12:52
pzhaopzhao
2401 silver badge8 bronze badges
answered May 27 at 12:52
pzhaopzhao
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2401 silver badge8 bronze badges
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add a comment
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A different base R possibility could be:
sapply(strsplit(confint, " ; ", fixed = TRUE), function(x) gsub("[^0-9.-]+", "\1", x) [1])
sapply(strsplit(confint, " ; ", fixed = TRUE), function(x) gsub("[^0-9.-]+", "\1", x) [2])
[1] "0.741" "263.917" "12.788" "0.680" "0.650" "0.719"
[1] "2.233" "402.154" "17.975" "2.450" "1.827" "2.190"
If you need it as a numeric vector:
sapply(strsplit(confint, " ; ", fixed = TRUE), function(x) as.numeric(gsub("[^0-9.-]+", "\1", x)) [1])
sapply(strsplit(confint, " ; ", fixed = TRUE), function(x) as.numeric(gsub("[^0-9.-]+", "\1", x)) [2])
answered May 27 at 12:49
tmfmnktmfmnk
12.4k1 gold badge10 silver badges26 bronze badges
@Ronak Shah I updated my post according some of your ideas. Thank you :)
– tmfmnk
May 27 at 12:59
I realised it needed some more tweaking to get exact output but I think it was a similar approach. +1 to you.
– Ronak Shah
May 27 at 13:21
add a comment
|
A different base R possibility could be:
sapply(strsplit(confint, " ; ", fixed = TRUE), function(x) gsub("[^0-9.-]+", "\1", x) [1])
sapply(strsplit(confint, " ; ", fixed = TRUE), function(x) gsub("[^0-9.-]+", "\1", x) [2])
[1] "0.741" "263.917" "12.788" "0.680" "0.650" "0.719"
[1] "2.233" "402.154" "17.975" "2.450" "1.827" "2.190"
If you need it as a numeric vector:
sapply(strsplit(confint, " ; ", fixed = TRUE), function(x) as.numeric(gsub("[^0-9.-]+", "\1", x)) [1])
sapply(strsplit(confint, " ; ", fixed = TRUE), function(x) as.numeric(gsub("[^0-9.-]+", "\1", x)) [2])
answered May 27 at 12:49
tmfmnktmfmnk
12.4k1 gold badge10 silver badges26 bronze badges
@Ronak Shah I updated my post according some of your ideas. Thank you :)
– tmfmnk
May 27 at 12:59
I realised it needed some more tweaking to get exact output but I think it was a similar approach. +1 to you.
– Ronak Shah
May 27 at 13:21
add a comment
|
A different base R possibility could be:
sapply(strsplit(confint, " ; ", fixed = TRUE), function(x) gsub("[^0-9.-]+", "\1", x) [1])
sapply(strsplit(confint, " ; ", fixed = TRUE), function(x) gsub("[^0-9.-]+", "\1", x) [2])
[1] "0.741" "263.917" "12.788" "0.680" "0.650" "0.719"
[1] "2.233" "402.154" "17.975" "2.450" "1.827" "2.190"
If you need it as a numeric vector:
sapply(strsplit(confint, " ; ", fixed = TRUE), function(x) as.numeric(gsub("[^0-9.-]+", "\1", x)) [1])
sapply(strsplit(confint, " ; ", fixed = TRUE), function(x) as.numeric(gsub("[^0-9.-]+", "\1", x)) [2])
answered May 27 at 12:49
tmfmnktmfmnk
12.4k1 gold badge10 silver badges26 bronze badges
A different base R possibility could be:
sapply(strsplit(confint, " ; ", fixed = TRUE), function(x) gsub("[^0-9.-]+", "\1", x) [1])
sapply(strsplit(confint, " ; ", fixed = TRUE), function(x) gsub("[^0-9.-]+", "\1", x) [2])
[1] "0.741" "263.917" "12.788" "0.680" "0.650" "0.719"
[1] "2.233" "402.154" "17.975" "2.450" "1.827" "2.190"
If you need it as a numeric vector:
sapply(strsplit(confint, " ; ", fixed = TRUE), function(x) as.numeric(gsub("[^0-9.-]+", "\1", x)) [1])
sapply(strsplit(confint, " ; ", fixed = TRUE), function(x) as.numeric(gsub("[^0-9.-]+", "\1", x)) [2])
answered May 27 at 12:49
tmfmnktmfmnk
12.4k1 gold badge10 silver badges26 bronze badges
edited May 27 at 12:58
answered May 27 at 12:49
tmfmnktmfmnk
12.4k1 gold badge10 silver badges26 bronze badges
answered May 27 at 12:49
tmfmnktmfmnk
12.4k1 gold badge10 silver badges26 bronze badges
answered May 27 at 12:49
tmfmnktmfmnk
12.4k1 gold badge10 silver badges26 bronze badges
12.4k1 gold badge10 silver badges26 bronze badges
@Ronak Shah I updated my post according some of your ideas. Thank you :)
– tmfmnk
May 27 at 12:59
I realised it needed some more tweaking to get exact output but I think it was a similar approach. +1 to you.
– Ronak Shah
May 27 at 13:21
add a comment
|
@Ronak Shah I updated my post according some of your ideas. Thank you :)
– tmfmnk
May 27 at 12:59
I realised it needed some more tweaking to get exact output but I think it was a similar approach. +1 to you.
– Ronak Shah
May 27 at 13:21
@Ronak Shah I updated my post according some of your ideas. Thank you :)
– tmfmnk
May 27 at 12:59
@Ronak Shah I updated my post according some of your ideas. Thank you :)
– tmfmnk
May 27 at 12:59
I realised it needed some more tweaking to get exact output but I think it was a similar approach. +1 to you.
– Ronak Shah
May 27 at 13:21
I realised it needed some more tweaking to get exact output but I think it was a similar approach. +1 to you.
– Ronak Shah
May 27 at 13:21
add a comment
|
Meanwhile, I came up with another base R solution
lower <- as.numeric(sub(".]*", "", sub(";.*", "", confint)))
upper <- as.numeric(sub("].*", "", sub(".*;", "", confint)))
Thank you all!
answered May 27 at 13:22
FatetaFateta
1298 bronze badges
add a comment
|
Meanwhile, I came up with another base R solution
lower <- as.numeric(sub(".]*", "", sub(";.*", "", confint)))
upper <- as.numeric(sub("].*", "", sub(".*;", "", confint)))
Thank you all!
answered May 27 at 13:22
FatetaFateta
1298 bronze badges
add a comment
|
Meanwhile, I came up with another base R solution
lower <- as.numeric(sub(".]*", "", sub(";.*", "", confint)))
upper <- as.numeric(sub("].*", "", sub(".*;", "", confint)))
Thank you all!
answered May 27 at 13:22
FatetaFateta
1298 bronze badges
Meanwhile, I came up with another base R solution
lower <- as.numeric(sub(".]*", "", sub(";.*", "", confint)))
upper <- as.numeric(sub("].*", "", sub(".*;", "", confint)))
Thank you all!
answered May 27 at 13:22
FatetaFateta
1298 bronze badges
edited Jun 11 at 5:31
answered May 27 at 13:22
FatetaFateta
1298 bronze badges
answered May 27 at 13:22
FatetaFateta
1298 bronze badges
answered May 27 at 13:22
FatetaFateta
1298 bronze badges
1298 bronze badges
add a comment
|
add a comment
|
You can use functions from the stringr library.
You can split strings with str_split() according a specific character (; in your case), then remove character with str_remove() ([ and ] in your case) and you will obtain what you want.
str_remove(str_split_fixed(confint, ";", n = 2)[,1], '\[') %>% as.numeric()
# [1] 0.741 263.917 12.788 0.680 0.650 0.719
answered May 27 at 12:38
demarsylvaindemarsylvain
1,5132 gold badges7 silver badges21 bronze badges
add a comment
|
You can use functions from the stringr library.
You can split strings with str_split() according a specific character (; in your case), then remove character with str_remove() ([ and ] in your case) and you will obtain what you want.
str_remove(str_split_fixed(confint, ";", n = 2)[,1], '\[') %>% as.numeric()
# [1] 0.741 263.917 12.788 0.680 0.650 0.719
answered May 27 at 12:38
demarsylvaindemarsylvain
1,5132 gold badges7 silver badges21 bronze badges
add a comment
|
You can use functions from the stringr library.
You can split strings with str_split() according a specific character (; in your case), then remove character with str_remove() ([ and ] in your case) and you will obtain what you want.
str_remove(str_split_fixed(confint, ";", n = 2)[,1], '\[') %>% as.numeric()
# [1] 0.741 263.917 12.788 0.680 0.650 0.719
answered May 27 at 12:38
demarsylvaindemarsylvain
1,5132 gold badges7 silver badges21 bronze badges
You can use functions from the stringr library.
You can split strings with str_split() according a specific character (; in your case), then remove character with str_remove() ([ and ] in your case) and you will obtain what you want.
str_remove(str_split_fixed(confint, ";", n = 2)[,1], '\[') %>% as.numeric()
# [1] 0.741 263.917 12.788 0.680 0.650 0.719
answered May 27 at 12:38
demarsylvaindemarsylvain
1,5132 gold badges7 silver badges21 bronze badges
edited May 27 at 12:45
answered May 27 at 12:38
demarsylvaindemarsylvain
1,5132 gold badges7 silver badges21 bronze badges
answered May 27 at 12:38
demarsylvaindemarsylvain
1,5132 gold badges7 silver badges21 bronze badges
answered May 27 at 12:38
demarsylvaindemarsylvain
1,5132 gold badges7 silver badges21 bronze badges
1,5132 gold badges7 silver badges21 bronze badges
add a comment
|
add a comment
|
A tidyverse solution:
library(dplyr)
library(tidyr)
df = data.frame(confint)
df = df %>%
mutate(confint = gsub("(\[|\])","",confint)) %>%
separate(confint,c("lower","upper"),";",convert=T)
answered May 27 at 14:12
FinoFino
1,2405 silver badges16 bronze badges
add a comment
|
A tidyverse solution:
library(dplyr)
library(tidyr)
df = data.frame(confint)
df = df %>%
mutate(confint = gsub("(\[|\])","",confint)) %>%
separate(confint,c("lower","upper"),";",convert=T)
answered May 27 at 14:12
FinoFino
1,2405 silver badges16 bronze badges
add a comment
|
A tidyverse solution:
library(dplyr)
library(tidyr)
df = data.frame(confint)
df = df %>%
mutate(confint = gsub("(\[|\])","",confint)) %>%
separate(confint,c("lower","upper"),";",convert=T)
answered May 27 at 14:12
FinoFino
1,2405 silver badges16 bronze badges
A tidyverse solution:
library(dplyr)
library(tidyr)
df = data.frame(confint)
df = df %>%
mutate(confint = gsub("(\[|\])","",confint)) %>%
separate(confint,c("lower","upper"),";",convert=T)
answered May 27 at 14:12
FinoFino
1,2405 silver badges16 bronze badges
answered May 27 at 14:12
FinoFino
1,2405 silver badges16 bronze badges
answered May 27 at 14:12
FinoFino
1,2405 silver badges16 bronze badges
answered May 27 at 14:12
FinoFino
1,2405 silver badges16 bronze badges
1,2405 silver badges16 bronze badges
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