Hi, how can i find an indice from an vetor to know his position.
ex: I have a vector non ordered with more than 1000 indices. I find the mean of this vector end want to know his position.
it seems to me likely that once you calculate the mean of the vector, you might find that none, or all of the members of the vector equal that mean value. What should be returned in such cases?
Generally though, the which() function is a fine approach to looking up an index in of a member of a vector. for example
my_vec <- c(1,100,99,101,100)
which(my_vec == 99) # answer 3
which(my_vec == 100) # answer 2 , only first found is returned
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Hi, I'm trying to get all number above the mean and above the indice of that mean. but its nor working. if you know how to do it better, i'm learning 
library(dslabs)
data("heights")
male <- heights$height[heights$sex=="Male"]
female <- heights$height[heights$sex=="Female"]
avg_male <- mean(male)
avg_female <- mean(female)
ind_male <- which(male == avg_male)
avg_female <- as.integer(avg_female)
ind_male <- which(male == avg_male) #not working
ind_female <- which(female==avh_female) #not working
abv_avg_male <- c(sort(male)[ind_male:length(male)]
abv_avg_female <- c(sort(female)[ind_female:length(female)]
abv_ind_male <- c(male[ind_male:length(male)]
abv_ind_female <- c(female[ind_female:length(female)]
I understand wanting to get all that are above the mean, but not sure what it means that you also want the indice be above ? given that you have unordered data... do you intend to first order your data so that those above the average will also be above the indice. ?
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No, I want to get the unordered data above the mean and ordered data above the mean too, to undertand how to do it. but in this case, the mean get two middle numbers and divide by 2. I doesnt know if I can do without we conditional expression
opsy, forgot what I said about the mean, I was thinking about the median. It will work better if was median. I guess
In the code here, I dont consider female case, you can adapt the male example to do that.
I also didnt do the full dataset, but a random sample of 20.
library(dslabs)
library(tidyverse )
set.seed(2)
options(pillar.sigfig=4) # i want to see 4 significant digits in my tibble print
# im going to work with a random 20 observations rather than the full dataset
# also having the dataframe in the tibble flavour for prettier printing
short_h <-sample_n( dslabs::heights, size = 20) %>% as_tibble()
#example of using glimpse to get a view of the dataframe on its side
glimpse(short_h)
#Observations: 20
#Variables: 2
$ sex <fct> Male, Male, Male, Male, Male, Male, Male, Male, Male, Male, Male, Female, Male, Fem...
$ height <dbl> 66.00000, 68.00000, 61.00000, 66.92000, 70.00000, 71.00000, 67.00000, 70.00000, 68....
male_vec<- short_h$height[short_h$sex=="Male"]
male_vec
[1] 66.00000 68.00000 61.00000 66.92000 70.00000 71.00000 67.00000 70.00000 68.00000 66.92913 66.14173
[12] 71.00000 65.00000 67.00000 67.00000 69.00000 73.00000 62.99213
avg_male <- median(male_vec)
avg_male
[1] 67
#while the median of 1,2,3 -- median(1:3) == 2 , the median of 1,2 ==1.5 therefore finding the first individual in the list that exactly matches would not be possible
# we have to apply some measure of distance
male_df <- short_h %>% filter(sex=="Male")
male_df
# A tibble: 18 x 2
sex height
<fct> <dbl>
1 Male 66
2 Male 68
3 Male 61
4 Male 66.92
5 Male 70
6 Male 71
7 Male 67
8 Male 70
9 Male 68
10 Male 66.93
11 Male 66.14
12 Male 71
13 Male 65
14 Male 67
15 Male 67
16 Male 69
17 Male 73
18 Male 62.99
male_df2 <- male_df %>% mutate(rnum = row_number(),
abs_dist_from_avg = abs(height-avg_male))
male_df2
# A tibble: 18 x 4
sex height rnum abs_dist_from_avg
<fct> <dbl> <int> <dbl>
1 Male 66 1 1
2 Male 68 2 1
3 Male 61 3 6
4 Male 66.92 4 0.08000
5 Male 70 5 3
6 Male 71 6 4
7 Male 67 7 0
8 Male 70 8 3
9 Male 68 9 1
10 Male 66.93 10 0.07087
11 Male 66.14 11 0.8583
12 Male 71 12 4
13 Male 65 13 2
14 Male 67 14 0
15 Male 67 15 0
16 Male 69 16 2
17 Male 73 17 6
18 Male 62.99 18 4.008
male_df3 <- arrange(male_df2,abs_dist_from_avg,rnum)
male_df3
# A tibble: 18 x 4
sex height rnum abs_dist_from_avg
<fct> <dbl> <int> <dbl>
1 Male 67 7 0
2 Male 67 14 0
3 Male 67 15 0
4 Male 66.93 10 0.07087
5 Male 66.92 4 0.08000
6 Male 66.14 11 0.8583
7 Male 66 1 1
8 Male 68 2 1
9 Male 68 9 1
10 Male 65 13 2
11 Male 69 16 2
12 Male 70 5 3
13 Male 70 8 3
14 Male 71 6 4
15 Male 71 12 4
16 Male 62.99 18 4.008
17 Male 61 3 6
18 Male 73 17 6
male_df4 <- head(male_df3,1) # take the top row
male_df4
# A tibble: 1 x 4
sex height rnum abs_dist_from_avg
<fct> <dbl> <int> <dbl>
1 Male 67 7 0
avg_male_position <- pull(male_df4,rnum) # take the single indice/row number from it (rnum)
avg_male_position
[1] 7
# we can go back to male_df2 and add our final metrics to that
final_male_df <- male_df2 %>% mutate(height_ge_median = height >= avg_male,
row_ge_avg_pos = rnum >= avg_male_position)
final_male_df
# A tibble: 18 x 6
sex height rnum abs_dist_from_avg height_ge_median row_ge_avg_pos
<fct> <dbl> <int> <dbl> <lgl> <lgl>
1 Male 66 1 1 FALSE FALSE
2 Male 68 2 1 TRUE FALSE
3 Male 61 3 6 FALSE FALSE
4 Male 66.92 4 0.08000 FALSE FALSE
5 Male 70 5 3 TRUE FALSE
6 Male 71 6 4 TRUE FALSE
7 Male 67 7 0 TRUE TRUE
8 Male 70 8 3 TRUE TRUE
9 Male 68 9 1 TRUE TRUE
10 Male 66.93 10 0.07087 FALSE TRUE
11 Male 66.14 11 0.8583 FALSE TRUE
12 Male 71 12 4 TRUE TRUE
13 Male 65 13 2 FALSE TRUE
14 Male 67 14 0 TRUE TRUE
15 Male 67 15 0 TRUE TRUE
16 Male 69 16 2 TRUE TRUE
17 Male 73 17 6 TRUE TRUE
18 Male 62.99 18 4.008 FALSE TRUEThis topic was automatically closed 7 days after the last reply. New replies are no longer allowed.