将TukeyHSD比较（第一列）分离成新的数据框架。将一列拆分为两列。

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英文:

Separate out TukeyHSD comparisons (first column) into new data frame. Split one column into two

问题

I understand that you want to separate the first column of your data into two new columns in a data frame, where the values like 0.1-0 should be split into 0.1 and 0. Here's how you can do it in R:

  1. # Your original data
  2. data <- read.table(text = "
  3.                    diff        lwr       upr     p adj
  4. 0.1-0          -7.6925867  -79.28269  63.89752 0.9999650
  5. 1-0             9.6882972  -40.93355  60.31014 0.9984060
  6. 10-0            4.1546350  -46.46721  54.77648 0.9999945
  7. 100-0           7.4767801  -44.70303  59.65659 0.9997580
  8. 1000-0          7.9958523  -41.34422  57.33593 0.9994555
  9. 10000-0        11.6970426  -42.41999  65.81407 0.9965756
  10. 100000-0     -104.3276885 -158.44472 -50.21066 0.0000106
  11. 1-0.1          17.3808839  -54.20922  88.97099 0.9930363
  12. 10-0.1         11.8472217  -59.74288  83.43732 0.9993756
  13. 100-0.1        15.1693668  -57.53073  87.86946 0.9972505
  14. 1000-0.1       15.6884389  -55.00112  86.37800 0.9959677
  15. 10000-0.1      19.3896293  -54.71317  93.49243 0.9891406
  16. 100000-0.1    -96.6351019 -170.73790 -22.53230 0.0038992
  17. 10-1           -5.5336622  -56.15551  45.08819 0.9999607
  18. 100-1          -2.2115171  -54.39132  49.96829 0.9999999
  19. 1000-1         -1.6924450  -51.03252  47.64763 1.0000000
  20. 10000-1         2.0087454  -52.10829  56.12578 1.0000000
  21. 100000-1     -114.0159857 -168.13302 -59.89895 0.0000019
  22. 100-10          3.3221451  -48.85766  55.50195 0.9999990
  23. 1000-10         3.8412172  -45.49886  53.18129 0.9999962
  24. 10000-10        7.5424076  -46.57462  61.65944 0.9997987
  25. 100000-10    -108.4823236 -162.59935 -54.36529 0.0000051
  26. 1000-100        0.5190721  -50.41818  51.45632 1.0000000
  27. 10000-100       4.2202625  -51.35684  59.79736 0.9999968
  28. 100000-100   -111.8044686 -167.38157 -56.22737 0.0000047
  29. 10000-1000      3.7011904  -49.21879  56.62117 0.9999982
  30. 100000-1000  -112.3235408 -165.24352 -59.40356 0.0000016
  31. 100000-10000 -116.0247312 -173.42451 -58.62495 0.0000043
  32. ", header = TRUE, row.names = 1, stringsAsFactors = FALSE)
  34. # Create a new data frame
  35. new_data <- data.frame(Group1 = as.numeric(sub('-.*', '', row.names(data))),
  36.                        Group2 = as.numeric(sub('.*-', '', row.names(data))),
  37.                        P = data$p.adj)
  39. # Print the new data frame
  40. print(new_data)

This code will create a new data frame new_data with the columns Group1, Group2, and P, where Group1 contains the values before the hyphen -, Group2 contains the values after the hyphen -, and P contains the p.adj values from your original data.

英文:

I am performing a TwoWay ANOVA on some data. When performing the TukeyHSD post-hoc, the following output is given:

  1.                    diff        lwr       upr     p adj
  2. 0.1-0          -7.6925867  -79.28269  63.89752 0.9999650
  3. 1-0             9.6882972  -40.93355  60.31014 0.9984060
  4. 10-0            4.1546350  -46.46721  54.77648 0.9999945
  5. 100-0           7.4767801  -44.70303  59.65659 0.9997580
  6. 1000-0          7.9958523  -41.34422  57.33593 0.9994555
  7. 10000-0        11.6970426  -42.41999  65.81407 0.9965756
  8. 100000-0     -104.3276885 -158.44472 -50.21066 0.0000106
  9. 1-0.1          17.3808839  -54.20922  88.97099 0.9930363
  10. 10-0.1         11.8472217  -59.74288  83.43732 0.9993756
  11. 100-0.1        15.1693668  -57.53073  87.86946 0.9972505
  12. 1000-0.1       15.6884389  -55.00112  86.37800 0.9959677
  13. 10000-0.1      19.3896293  -54.71317  93.49243 0.9891406
  14. 100000-0.1    -96.6351019 -170.73790 -22.53230 0.0038992
  15. 10-1           -5.5336622  -56.15551  45.08819 0.9999607
  16. 100-1          -2.2115171  -54.39132  49.96829 0.9999999
  17. 1000-1         -1.6924450  -51.03252  47.64763 1.0000000
  18. 10000-1         2.0087454  -52.10829  56.12578 1.0000000
  19. 100000-1     -114.0159857 -168.13302 -59.89895 0.0000019
  20. 100-10          3.3221451  -48.85766  55.50195 0.9999990
  21. 1000-10         3.8412172  -45.49886  53.18129 0.9999962
  22. 10000-10        7.5424076  -46.57462  61.65944 0.9997987
  23. 100000-10    -108.4823236 -162.59935 -54.36529 0.0000051
  24. 1000-100        0.5190721  -50.41818  51.45632 1.0000000
  25. 10000-100       4.2202625  -51.35684  59.79736 0.9999968
  26. 100000-100   -111.8044686 -167.38157 -56.22737 0.0000047
  27. 10000-1000      3.7011904  -49.21879  56.62117 0.9999982
  28. 100000-1000  -112.3235408 -165.24352 -59.40356 0.0000016
  29. 100000-10000 -116.0247312 -173.42451 -58.62495 0.0000043

Is there an easy way to separate out the first column into a new data frame, (this is to be the data input for p_stat_manual in a ggplot, which requires the dataframe to be Group 1 | Group 2 | P))

i.e. column 1, row 1 says: 0.1-0. In the new dataframe how can I have 0.1 in column Group1 and 0 in column Group2 etc...

Currently I am writing by hand as follows, but this is time-consuming and is easy to introduce errors:

  1. group1 &lt;- (c(1, 10, 100, 1000, 10000, 100000, &quot;DMSO&quot;, 10, 100, 1000, 10000, 100000, &quot;DMSO&quot;, 100, 1000, 10000, 100000, &quot;DMSO&quot;, 1000, 10000, 100000, &quot;DMSO&quot;, 10000, 100000, &quot;DMSO&quot;, 100000, &quot;DMSO&quot;, &quot;DMSO&quot;))
  2.  
  3. group2 &lt;- factor(c(0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 10, 10, 10, 10, 10, 100, 100, 100, 100, 1000, 1000,1000, 10000, 10000, 100000))

答案1

得分: 2

这里是如何执行的示例，我使用了 ?TukeyHSD 的示例。

  1. library(dplyr)
  2. library(tibble)
  3. library(tidyr)
  5. fm1 <- aov(breaks ~ wool + tension, data = warpbreaks)
  7. tukey <- TukeyHSD(fm1, "tension", ordered = TRUE)
  9. tukey$tension %>%
  10.   as.data.frame() %>%
  11.   rownames_to_column() %>%
  12.   separate(col = rowname, into = c("var1", "var2"))
英文:

Here an example of how to do it, I used the ?TukeyHSD example.

  1. library(dplyr)
  2. library(tibble)
  3. library(tidyr)
  5. fm1 &lt;- aov(breaks ~ wool + tension, data = warpbreaks)
  7. tukey &lt;- TukeyHSD(fm1, &quot;tension&quot;, ordered = TRUE)
  9. tukey$tension %&gt;% 
  10.   as.data.frame() %&gt;% 
  11.   rownames_to_column() %&gt;% 
  12.   separate(col = rowname,into = c(&quot;var1&quot;,&quot;var2&quot;))

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  • 本文由 发表于 2023年1月9日 09:12:46
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