I want to develop a rule to extract certain rows from a matrix. I set up the example as follows:
mat1 = data.frame(matrix(nrow=508, ncol =5))
mat1[1:20,1] = rep(1,20)
mat1[1:20,2:5] = rnorm(20*4,0,1)
mat2 = data.frame(matrix(nrow=508, ncol =5))
seq1 <- seq(1,3,1)
mat2[1:27,1] = rep(seq1,9)
mat2[1:27,2:5] = rnorm(27*4,0,1)
mat3 = data.frame(matrix(nrow=508, ncol =5))
mat3[1:32,1] = rep(seq(1,4,1),8)
mat3[1:32,2:5] = rnorm(32*4,0,1)
colnames(mat1) = colnames(mat2) = colnames(mat3) = c("Cohort Number", "Alpha(t-1)", "date1", "date2", "date3")
mat.list <- list(mat1,mat2,mat3)
Example matrix
Cohort Number Alpha(t-1) date1 date2 date3
1 1 -1.76745451 -1.3227308 2.7099501 -0.13797329
2 1 -0.72651808 -0.8714317 1.3200554 0.76964663
3 1 -0.50325892 0.0742336 -0.6460628 0.30148135
4 1 0.79592650 0.1353875 -0.5694022 -0.59019913
5 1 1.94064961 0.2255595 0.3156252 -0.90996475
6 1 0.27134932 0.3966957 -1.9198976 0.23998928
7 1 -1.13272507 -0.8603225 -1.2042036 0.06609958
8 1 -2.12392748 1.0905405 -0.3788234 0.92850110
9 1 0.22038996 0.4500683 -1.4617004 0.58498275
10 1 0.26348734 -0.8340913 1.2631368 -1.48490518
11 1 0.26931077 -0.5230622 -0.6615288 1.45668453
12 1 -2.03067695 -0.6432484 0.4801026 0.01808834
13 1 1.25915656 -0.1116544 -0.3004298 -1.04072722
14 1 -2.27894271 -2.1058424 -0.3351053 -1.04132045
15 1 0.47742052 2.1564274 -0.4733351 -0.53152019
16 1 -1.57680089 -0.1340645 -0.3134633 0.53223567
17 1 0.25245813 -0.8243152 0.5998211 -1.01892301
18 1 0.18391447 -1.3500645 1.6059798 1.43359399
19 1 -0.09602031 1.4921338 -0.6455687 0.66385823
20 1 -0.13613759 2.2474816 0.7311762 -2.46849071
mat2[1:27,]
Cohort Number Alpha(t-1) date1 date2 date3
1 1 -0.76033920 1.317636591 -0.09684526 -0.08796725
2 2 0.05123185 -0.731591674 -0.37247406 0.04470346
3 3 -0.78460201 0.890336570 1.26737475 -0.39062992
4 1 -0.14111920 1.255008475 -0.32799815 -0.77277716
5 2 -0.46044451 1.175157970 0.82187906 0.54326905
6 3 -0.46804365 0.704203273 -2.04539007 -1.74782065
7 1 0.42009824 0.488807461 3.21093186 -0.13745029
8 2 1.27083389 -1.316989452 0.43565921 0.07870330
9 3 -0.16581119 1.872955624 -0.22399155 -0.79334562
10 1 -1.33436656 0.589311311 -1.03871415 -1.06221057
11 2 1.56584985 0.020699064 0.45691456 0.15858065
12 3 1.07756426 -0.045200151 0.05124461 -1.86633279
13 1 -1.01264994 -0.229406681 1.24954420 0.88846407
14 2 -0.09950713 -0.515798138 1.62560454 -0.20191909
15 3 -0.28319479 0.450854419 1.42963386 -1.11964154
16 1 0.51771608 -1.407248379 0.62626313 0.97775246
17 2 -0.43951262 -0.368739441 0.66564013 -0.79980882
18 3 -0.15865277 -0.231475146 0.37582330 0.93685867
19 1 -0.57758129 0.235550070 0.42480442 -0.14379249
20 2 -0.81726414 -1.207593079 -0.30000514 0.68967230
21 3 -0.72926703 -0.458849409 1.51162785 1.40921409
22 1 -0.32220454 0.334996561 1.26073381 -2.03405958
23 2 -0.51450039 -0.305634241 1.51021957 0.39775430
24 3 1.15476297 -1.040126709 -0.36192432 -0.37346894
25 1 -0.88053587 -0.006829769 -0.89855797 -0.39840858
26 2 -0.64435448 0.209561006 -0.13986834 -0.61308957
27 3 1.22492942 0.812693992 -1.32371617 -1.21852365
and
> mat3[1:32,]
Cohort Number Alpha(t-1) date1 date2 date3
1 1 -0.7657871 -0.35390862 -0.23539987 -1.8365309
2 2 -0.6631690 1.36450837 0.78403072 -0.8344993
3 3 -1.0134022 -0.28380021 0.72149463 -0.7890273
4 4 2.6419455 0.26998803 2.03606725 0.8099134
5 1 -0.1383910 0.90845134 1.09273919 0.4651443
6 2 -0.7549340 -0.23185551 2.21119705 -0.1386960
7 3 0.7296121 -1.09145187 -1.18092505 0.1510642
8 4 -0.5583415 0.71988405 0.09454476 -0.8661514
9 1 -0.2420894 -0.03215026 -2.51249946 1.1659027
10 2 -0.6434337 -0.13910557 -1.10373674 1.2377968
11 3 -0.6297123 2.09797419 0.87128407 -0.1351845
12 4 0.6674166 0.48707847 0.36373509 1.0680623
13 1 0.6254708 -0.61311671 0.82542494 1.7320687
14 2 -2.4704173 0.98460064 -1.10416042 2.9627952
15 3 -0.2544887 0.63177246 -0.39138717 1.6942072
16 4 -0.9807623 1.11882794 -0.47669974 1.2383798
17 1 -0.6900549 1.68086482 -0.01405476 -1.3099288
18 2 1.4510505 -0.04752782 1.49735258 0.2963673
19 3 -1.1355194 -1.76263532 -1.49318214 1.3524114
20 4 0.7168833 -0.76833639 0.60752304 -1.0647885
21 1 2.0004745 2.13931057 -1.35036048 -0.7694501
22 2 2.0985591 0.01569677 0.33975952 -1.4979973
23 3 0.1703261 -1.47625208 -1.13228671 0.5686501
24 4 0.2632233 -0.55672667 0.33428217 0.5341078
25 1 -0.2741324 -1.61301237 0.78861248 0.4982554
26 2 -0.8793897 -1.07266362 -0.78158128 0.9127354
27 3 0.3920579 -0.59869834 -0.76775259 1.8137107
28 4 -1.4088488 -0.54954542 0.32421016 0.7284813
29 1 -1.2421837 0.50599077 1.62464999 0.6801672
30 2 -2.8980422 0.42197236 0.45243582 1.4939070
31 3 0.3965108 -1.35877353 1.52230797 -1.6552039
32 4 0.8112229 0.51970084 0.30830797 -2.0563928
What I want to do:
For every matrix in mat.list I want to extract 6 rows of data, according to certain criteria, and place these rows as a data.frame in a list labelled Output1. I want to store all remaining rows as a data.frame in Output2.
The process:
1) Group data by cohort number.
2a. If there is 1 group (Cohort Number can only = 1). Move to column 2 and extract the 6 rows of matrix with the highest value for "Alpha(t-1)". Store these rows as a data.frame in a list named "Output1". Store all remaining rows as a data.frame in a list named "Output2".
2b. If there are 2 groups (Cohort number can = 1 or Cohort Number can =2) move to column 2 and extract the 3 rows with the largest "Alpha(t-1)" corresponding to Cohort Number ==1 and extract the 3 rows with largest"Alpha(t-1)" corresponding to Cohort Number == 2. Place the 6 rows extracted as a data.frame in a list named "Output1". Place all remaining rows as a data.frame in a list named "Output2".
2c. If there are 3 groups ("Cohort Number can = 1, Cohort Number can =2, Cohort Number can =3 ) move to column 2 and extract the 2 rows with the largest "Alpha(t-1)" corresponding to Cohort Number ==1, extract the 2 rows with the largest "Alpha(t-1)" corresponding to Cohort Number =2 and extract the 2 rows with the largest "Alpha(t-1)" corresponding to Cohort Number =3
2d. If there are 4 groups ("Cohort Number can = 1, Cohort Number can =2, Cohort Number can =3, Cohort Number = 4) move to column 2. Extract the 2 rows with the largest "Alpha(t-1)" corresponding to Cohort Number ==1. Extract the 2 row with the largest "Alpha(t-1)" corresponding to Cohort Number ==2. Extract the 1 row with the largest "Alpha(t-1)" corresponding to Cohort Number ==3 and Extract the 1 row with the largest "Alpha(t-1)" corresponding to Cohort Number ==4. Store the 6 key rows as a data.frame in Output1. Store all remaining rows as a data.frame in the list Output2.
Desired Output:
Output1 <- c()
Output2 <- c()
Output1[[1]] = mat1 %>% group_by(`Cohort Number`) %>% top_n(6, `Alpha(t-1)`)
Output1[[2]] = mat2 %>% group_by(`Cohort Number`) %>% top_n(2, `Alpha(t-1)`)
> Output1[[1]]
# A tibble: 6 x 5
# Groups: Cohort Number [1]
`Cohort Number` `Alpha(t-1)` date1 date2 date3
<dbl> <dbl> <dbl> <dbl> <dbl>
1 1 0.796 0.135 -0.569 -0.590
2 1 1.94 0.226 0.316 -0.910
3 1 0.271 0.397 -1.92 0.240
4 1 0.269 -0.523 -0.662 1.46
5 1 1.26 -0.112 -0.300 -1.04
6 1 0.477 2.16 -0.473 -0.532
> Output1[[2]]
# A tibble: 6 x 5
# Groups: Cohort Number [3]
`Cohort Number` `Alpha(t-1)` date1 date2 date3
<dbl> <dbl> <dbl> <dbl> <dbl>
1 1 0.420 0.489 3.21 -0.137
2 2 1.27 -1.32 0.436 0.0787
3 2 1.57 0.0207 0.457 0.159
4 1 0.518 -1.41 0.626 0.978
5 3 1.15 -1.04 -0.362 -0.373
6 3 1.22 0.813 -1.32 -1.22
Overall I need a function to do this because i have over 1000 matrices in my actual application and can't do this manually.
We can count the number of distinct values in Cohort Number and based on that select the value of n in top_n. For distinct values which are more than 3, we create vector of values to select in top_n for each Cohort Number.
library(tidyverse)
output1 <- map(mat.list, function(x) {
dist <- n_distinct(x$`Cohort Number`, na.rm = TRUE)
if(dist <= 3)
x %>%
group_by(`Cohort Number`) %>%
top_n(6/dist, `Alpha(t-1)`)
else
map2_df(list(2, 2, 1, 1),x %>% na.omit %>% group_split(`Cohort Number`),
~.y %>% top_n(.x, `Alpha(t-1)`))
})
and for output2, we use map2 with ant_join
output2 <- map2(mat.list, output1, anti_join)
Confirming the output
map_dbl(output1, nrow)
#[1] 6 6 6
map_dbl(output2, nrow)
#[1] 502 502 502
Related
I have the following dataframe, which is the result of a cluster analysis with ten 7-likert attitude scales for specific product benefits (see 'variable' column). At this, n is the number of persons stating a specific value for each Benefit and sum is the total sum of persons for each cluster. n2 is just the relative share of answers to all answers per cluster (n2=n/cum*100, which is basically %).
Now, I want to create a new column, aggregating / summing up the top-n (indicated in 'value' column) percent (indicated in n2) for each benefit, e.g. a new column "Top-3-Box" with e.g. a value of 46.5 for rows 1-7/Benefit.1 (which is the sum of the n2 of the rows with the top-3 value 7,6,5). It would be great if there would be a solution for this, which is instantly applicable in dplyr.
Please see the dataframe below:
cluster variable value n cum n2
<int> <chr> <dbl> <int> <int> <dbl>
1 1 Benefit.1 1 11 86 12.8
2 1 Benefit.1 2 11 86 12.8
3 1 Benefit.1 3 6 86 7
4 1 Benefit.1 4 18 86 20.9
5 1 Benefit.1 5 16 86 18.6
6 1 Benefit.1 6 14 86 16.3
7 1 Benefit.1 7 10 86 11.6
8 1 Benefit.10 1 10 86 11.6
9 1 Benefit.10 2 13 86 15.1
10 1 Benefit.10 3 8 86 9.3
# ... with 40 more rows
I highly appreciate your support!
We can do a group by sum of 'n2' by subsetting the values corresponding to the first 3 'value'
library(dplyr)
df1 %>%
group_by(cluster, variable) %>%
mutate(percent = sum(n2[value %in% 1:3]))
If the 'value' is already ordered per 'cluster', 'variable', then we can just subset the 'n2'
df1 %>%
group_by(cluster, variable) %>%
mutate(percent = sum(n2[1:3]))
I have this dataframe
r2 distance
1 33.64 67866
2 8.50 77229
3 15.07 109119
4 24.35 142279
5 7.74 143393
6 8.21 177670
7 12.26 216440
8 12.66 253751
9 26.31 282556
10 39.08 320816
I need to calculate the mean of column r2 for every block of rows where the distance between two values in the column distance is equal or less than 100000.
For this example the desired output would be:
mean_r2 diff_of_distance
1 17.86 75527 ## mean of rows 1 to 5; distance 5 - distance 1
2 13.91 66164 ## mean of rows 2 to 5; distance 5 - distance 2
3 13.84 68551 ## mean of rows 3 to 6; distance 6 - distance 3
4 13.14 74161 ## mean of rows 4 to 7; distance 7 - distance 4
5 9.40 73047 ## mean of rows 5 to 7; distance 7 - distance 5
6 11.04 76081 ## mean of rows 6 to 8; distance 8 - distance 6
and so on.
Edit 1: I have more than 100,000 rows.
Thanks.
Loop through each value of distance, minus this from the values in the distance vector and test if the result is less than 100000. This creates a boolean vector which you sum to identify the index at which the distance is greater than 100000 (i.e. bool becomes FALSE). Use this index to identify your block then take the mean of r2 in each block.
To speed up the code define your vector type and length (to avoid "growing vectors" on each iteration.
means <- vector("numeric", length = nrow(df))
rows <- vector("numeric", length = nrow(df))
distance_diff <- vector("numeric", length = nrow(df))
for (i in seq_along(df$distance)) {
dis_val <- df$distance[i] # the ith distance value
bools <- (df$distance - dis_val) < 100000 # bool indicating if difference between i and every value in vector is less than 100000
block_range <- sum(bools)# taking sum of bools identifies the value at which the distance becomes > 100000
rows[i] <- paste(as.character(i), "-", as.character(block_range))
means[i] <- mean(df$r2[i:block_range]) # take the mean of r2 in the range i to all rows where distance is < 100000
distance_diff[i] <- df$distance[block_range] - dis_val # minus the distance from the value before distance is > 100000 from i
}
data.frame(mean_r2 = means, rows= rows, diff_of_distance=distance_diff)
mean_r2 rows diff_of_distance
1 17.860000 1 - 5 75527
2 13.915000 2 - 5 66164
3 13.842500 3 - 6 68551
4 13.140000 4 - 7 74161
5 9.403333 5 - 7 73047
6 11.043333 6 - 8 76081
7 17.076667 7 - 9 66116
8 26.016667 8 - 10 67065
9 32.695000 9 - 10 38260
10 39.080000 10 - 10 0
You can try:
# your data
d <- read.table(text="r2 distance
1 33.64 67866
2 8.50 77229
3 15.07 109119
4 24.35 142279
5 7.74 143393
6 8.21 177670
7 12.26 216440
8 12.66 253751
9 26.31 282556
10 39.08 320816", header=T)
library(tidyverse) #dplyr_0.7.2
d %>%
mutate(index=1:n()) %>% add row index
group_by(index) %>% # group by this index
# calculate difference and find max row where diff < 100000
mutate(max_row=max(which(.$distance - distance < 100000, arr.ind=T))) %>%
# calculate mean
mutate(mean_r2=mean(.$r2[index:max_row])) %>%
# calculate the difference
mutate(diff_of_distance=.$distance[max_row] - .$distance[index]) %>%
# unite the columns
unite(rows, index, max_row, sep = "-")
# A tibble: 10 x 5
r2 distance rows mean_r2 diff_of_distance
* <dbl> <int> <chr> <dbl> <int>
1 33.64 67866 1-5 17.860000 75527
2 8.50 77229 2-5 13.915000 66164
3 15.07 109119 3-6 13.842500 68551
4 24.35 142279 4-7 13.140000 74161
5 7.74 143393 5-7 9.403333 73047
6 8.21 177670 6-8 11.043333 76081
7 12.26 216440 7-9 17.076667 66116
8 12.66 253751 8-10 26.016667 67065
9 26.31 282556 9-10 32.695000 38260
10 39.08 320816 10-10 39.080000 0
This works because group_by subsets the dataframe, thus you can access within mutate the respective distance value per group and calculate the difference with the complete vector using .$distance as this access the complete column regardless the group_by() function.
I am trying to order data by the column weightFisher. However, it is almost as if R does not process e values as low, because all the e values are skipped when I try to order from smallest to greatest.
Code:
resultTable_bon <- GenTable(GOdata_bon,
weightFisher = resultFisher_bon,
weightKS = resultKS_bon,
topNodes = 15136,
ranksOf = 'weightFisher'
)
head(resultTable_bon)
#create Fisher ordered df
indF <- order(resultTable_bon$weightFisher)
resultTable_bonF <- resultTable_bon[indF, ]
what resultTable_bon looks like:
GO.ID Term Annotated Significant Expected Rank in weightFisher
1 GO:0019373 epoxygenase P450 pathway 19 13 1.12 1
2 GO:0097267 omega-hydroxylase P450 pathway 9 7 0.53 2
3 GO:0042738 exogenous drug catabolic process 10 7 0.59 3
weightFisher weightKS
1 1.9e-12 0.79744
2 7.9e-08 0.96752
3 2.5e-07 0.96336
what "ordered" resultTable_bonF looks like:
GO.ID Term Annotated Significant Expected Rank in weightFisher
17 GO:0014075 response to amine 33 7 1.95 17
18 GO:0034372 very-low-density lipoprotein particle re... 11 5 0.65 18
19 GO:0060710 chorio-allantoic fusion 6 4 0.35 19
weightFisher weightKS
17 0.00014 0.96387
18 0.00016 0.83624
19 0.00016 0.92286
As #bhas says, it appears to be working precisely as you want it to. Maybe it's the use of head() that's confusing you?
To put your mind at ease, try it with something simpler
dtf <- data.frame(a=c(1, 8, 6, 2)^-10, b=c(7, 2, 1, 6))
dtf
# a b
# 1 1.000000e+00 7
# 2 9.313226e-10 2
# 3 1.653817e-08 1
# 4 9.765625e-04 6
dtf[order(dtf$a), ]
# a b
# 2 9.313226e-10 2
# 3 1.653817e-08 1
# 4 9.765625e-04 6
# 1 1.000000e+00 7
Try the following :
resultTable_bon$weightFisher <- as.numeric (resultTable_bon$weightFisher)
Then :
resultTable_bonF <- resultTable_bon[order(resultTable_bonF$weightFisher),]
I want to determine the length of the snow season in the following data frame:
DATE SNOW
1998-11-01 0
1998-11-02 0
1998-11-03 0.9
1998-11-04 1
1998-11-05 0
1998-11-06 1
1998-11-07 0.6
1998-11-08 1
1998-11-09 2
1998-11-10 2
1998-11-11 2.5
1998-11-12 3
1998-11-13 6.5
1999-01-01 15
1999-01-02 15
1999-01-03 19
1999-01-04 18
1999-01-05 17
1999-01-06 17
1999-01-07 17
1999-01-08 17
1999-01-09 16
1999-03-01 6
1999-03-02 5
1999-03-03 5
1999-03-04 5
1999-03-05 5
1999-03-06 2
1999-03-07 2
1999-03-08 1.6
1999-03-09 1.2
1999-03-10 1
1999-03-11 0.6
1999-03-12 0
1999-03-13 1
Snow season is defined by a snow depth (SNOW) of more than 1 cm for at least 10 consecutive days (so if there is snow one day in November but after it melts and depth is < 1 cm we consider the season not started).
My idea would be to determine:
1) the date of snowpack establishement (in my example 1998-11-08)
2) the date of "disappearing" (here 1999-03-11)
3) calculate the length of the period (nb of days between 1998-11-05 and 1999-03-11)
For the 3rd step I can easily get the number between 2 dates using this method.
But how to define the dates with conditions?
This is one way:
# copy data from clipboard
d <- read.table(text=readClipboard(), header=TRUE)
# coerce DATE to Date type, add event grouping variable that numbers the groups
# sequentially and has NA for values not in events.
d <- transform(d, DATE=as.Date(DATE),
event=with(rle(d$SNOW >= 1), rep(replace(ave(values, values, FUN=seq), !values, NA), lengths)))
# aggregate event lengths in days
event.days <- aggregate(DATE ~ event, data=d, function(x) as.numeric(max(x) - min(x), units='days'))
# get those events greater than 10 days
subset(event.days, DATE > 10)
# event DATE
# 3 3 122
You can also use the event grouping variable to find the start dates:
starts <- aggregate(DATE ~ event, data=d, FUN=head, 1)
# 1 1 1998-11-04
# 2 2 1998-11-06
# 3 3 1998-11-08
# 4 4 1999-03-13
And then merge this with event.days:
merge(event.days, starts, by='event')
# event DATE.x DATE.y
# 1 1 0 1998-11-04
# 2 2 0 1998-11-06
# 3 3 122 1998-11-08
# 4 4 0 1999-03-13
I have a data frame with 3 columns. a,b,c. There are multiple rows corresponding to each unique value of column a. I want to select top 5 rows corresponding to each unique value of column a. column c is some value and the data frame is already sorted by it in descending order, so that would not be a problem. Can anyone please suggest how can I do this in R.
Stealing #ptocquin's example, here's how you can use base function by. You can flatten the result using do.call (see below).
> by(data = data, INDICES = data$a, FUN = function(x) head(x, 5))
# or by(data = data, INDICES = data$a, FUN = head, 5)
data$a: 1
a b c
21 1 0.1188552 1.6389895
41 1 1.0182033 1.4811359
61 1 -0.8795879 0.7784072
81 1 0.6485745 0.7734652
31 1 1.5102255 0.7107957
------------------------------------------------------------
data$a: 2
a b c
15 2 -1.09704040 1.1710693
85 2 0.42914795 0.8826820
65 2 -1.01480957 0.6736782
45 2 -0.07982711 0.3693384
35 2 -0.67643885 -0.2170767
------------------------------------------------------------
A similar thing could be achieved by splitting your data.frame based on a and then using lapply to step through each element subsetting first n rows.
split.data <- split(data, data$a)
subsetted.data <- lapply(split.data, FUN = function(x) head(x, 5)) # or ..., FUN = head, 5) like above
flatten.data <- do.call("rbind", subsetted.data)
head(flatten.data)
a b c
1.21 1 0.11885516 1.63898947
1.41 1 1.01820329 1.48113594
1.61 1 -0.87958790 0.77840718
1.81 1 0.64857445 0.77346517
1.31 1 1.51022545 0.71079568
2.15 2 -1.09704040 1.17106930
2.85 2 0.42914795 0.88268205
2.65 2 -1.01480957 0.67367823
2.45 2 -0.07982711 0.36933837
2.35 2 -0.67643885 -0.21707668
Here is my try :
library(plyr)
data <- data.frame(a=rep(sample(1:20,10),10),b=rnorm(100),c=rnorm(100))
data <- data[rev(order(data$c)),]
head(data, 15)
a b c
28 6 1.69611039 1.720081
91 11 1.62656460 1.651574
70 9 -1.17808386 1.641954
6 15 1.23420550 1.603140
23 7 0.70854914 1.588352
51 11 -1.41234359 1.540738
19 10 2.83730734 1.522825
49 10 0.39313579 1.370831
80 9 -0.59445323 1.327825
59 10 -0.55538404 1.214901
18 6 0.08445888 1.152266
86 15 0.53027267 1.066034
69 10 -1.89077464 1.037447
62 1 -0.43599566 1.026505
3 7 0.78544009 1.014770
result <- ddply(data, .(a), "head", 5)
head(result, 15)
a b c
1 1 -0.43599566 1.02650544
2 1 -1.55113486 0.36380251
3 1 0.68608364 0.30911430
4 1 -0.85406406 0.05555500
5 1 -1.83894595 -0.11850847
6 5 -1.79715809 0.77760033
7 5 0.82814909 0.22401278
8 5 -1.52726859 0.06745849
9 5 0.51655092 -0.02737905
10 5 -0.44004646 -0.28106808
11 6 1.69611039 1.72008079
12 6 0.08445888 1.15226601
13 6 -1.99465060 0.82214319
14 6 0.43855489 0.76221979
15 6 -2.15251353 0.64417757