How to extract certain values from text file to find average of the group of numbers?

Question

Sujata Dhar 2021-9-10

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1450849-how-to-extract-certain-values-from-text-file-to-find-average-of-the-group-of-numbers

评论： Star Strider 2021-9-11

Can you please help me to extract certain column of numbers from a text file. I want to find their average. The following .txt file is as follows. I want only # 25 to 30 sigma values.

Number of unknowns: 1374

Number of observations: 28854 of 28854

Number of constraints: 1246

Number of datum stations: 14

Number of datum sources: 0

vTPv: 67100.6 +/- 1944.94

redundancy: 28726

chi^2: 1.5282 +/- 0.0220971

List of estimated parameters

.----------------------------------------------------------------------------------------------------.

|-------|-----------|----------|---------------------|-----------------------|-----------------------|

| 0 | PWL_CLK | AGGO | 2021.05.18 17:00:00 | 0.62888 [cm] | 1.99957 [cm] |

| 1 | PWL_CLK | AGGO | 2021.05.18 18:00:00 | 0.91947 [cm] | 4.93553 [cm] |

| 2 | PWL_CLK | AGGO | 2021.05.18 19:00:00 | 1.64927 [cm] | 8.35032 [cm] |

| 3 | PWL_CLK | AGGO | 2021.05.18 20:00:00 | 2.28501 [cm] | 10.70383 [cm] |

| 4 | PWL_CLK | AGGO | 2021.05.18 21:00:00 | 2.92039 [cm] | 12.11574 [cm] |

| 5 | PWL_CLK | AGGO | 2021.05.18 22:00:00 | 3.51718 [cm] | 12.86139 [cm] |

| 6 | PWL_CLK | AGGO | 2021.05.18 23:00:00 | 4.05227 [cm] | 12.85490 [cm] |

| 7 | PWL_CLK | AGGO | 2021.05.19 00:00:00 | 4.56631 [cm] | 12.34585 [cm] |

| 8 | PWL_CLK | AGGO | 2021.05.19 01:00:00 | 5.02033 [cm] | 11.29844 [cm] |

| 9 | PWL_CLK | AGGO | 2021.05.19 02:00:00 | 5.44759 [cm] | 10.22204 [cm] |

| 10 | PWL_CLK | AGGO | 2021.05.19 03:00:00 | 5.82947 [cm] | 9.16751 [cm] |

| 11 | PWL_CLK | AGGO | 2021.05.19 04:00:00 | 6.17638 [cm] | 8.30392 [cm] |

| 12 | PWL_CLK | AGGO | 2021.05.19 05:00:00 | 6.49624 [cm] | 7.89360 [cm] |

| 13 | PWL_CLK | AGGO | 2021.05.19 06:00:00 | 6.79364 [cm] | 7.92088 [cm] |

| 14 | PWL_CLK | AGGO | 2021.05.19 07:00:00 | 7.06446 [cm] | 8.59745 [cm] |

| 15 | PWL_CLK | AGGO | 2021.05.19 08:00:00 | 7.30998 [cm] | 9.59593 [cm] |

| 16 | PWL_CLK | AGGO | 2021.05.19 09:00:00 | 7.55144 [cm] | 10.44086 [cm] |

| 17 | PWL_CLK | AGGO | 2021.05.19 10:00:00 | 7.77601 [cm] | 11.36565 [cm] |

| 18 | PWL_CLK | AGGO | 2021.05.19 11:00:00 | 8.00764 [cm] | 11.76791 [cm] |

| 19 | PWL_CLK | AGGO | 2021.05.19 12:00:00 | 8.24088 [cm] | 11.82715 [cm] |

| 20 | PWL_CLK | AGGO | 2021.05.19 13:00:00 | 8.49287 [cm] | 11.09234 [cm] |

| 21 | PWL_CLK | AGGO | 2021.05.19 14:00:00 | 8.76043 [cm] | 9.66000 [cm] |

| 22 | PWL_CLK | AGGO | 2021.05.19 15:00:00 | 9.06284 [cm] | 7.28851 [cm] |

| 23 | PWL_CLK | AGGO | 2021.05.19 16:00:00 | 9.41047 [cm] | 4.15836 [cm] |

| 24 | PWL_CLK | AGGO | 2021.05.19 17:00:00 | 9.75290 [cm] | 1.99957 [cm] |

| 25 | PWL_CLK | HART15M | 2021.05.18 17:00:00 | 0.57717 [cm] | 1.75935 [cm] |

| 26 | PWL_CLK | HART15M | 2021.05.18 18:00:00 | 0.88342 [cm] | 4.97802 [cm] |

| 27 | PWL_CLK | HART15M | 2021.05.18 19:00:00 | 1.59244 [cm] | 8.59066 [cm] |

| 28 | PWL_CLK | HART15M | 2021.05.18 20:00:00 | 2.26089 [cm] | 10.78367 [cm] |

| 29 | PWL_CLK | HART15M | 2021.05.18 21:00:00 | 2.90160 [cm] | 12.28015 [cm] |

| 30 | PWL_CLK | HART15M | 2021.05.18 22:00:00 | 3.49047 [cm] | 13.06497 [cm] |

| 31 | PWL_CLK | HART15M | 2021.05.18 23:00:00 | 4.03382 [cm] | 13.11746 [cm] |

| 32 | PWL_CLK | HART15M | 2021.05.19 00:00:00 | 4.54147 [cm] | 12.73883 [cm] |

| 33 | PWL_CLK | HART15M | 2021.05.19 01:00:00 | 5.00205 [cm] | 11.97155 [cm] |

| 34 | PWL_CLK | HART15M | 2021.05.19 02:00:00 | 5.42749 [cm] | 10.89467 [cm] |

| 35 | PWL_CLK | HART15M | 2021.05.19 03:00:00 | 5.81582 [cm] | 9.73427 [cm] |

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Answer 1

Star Strider 2021-9-10

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1450849-how-to-extract-certain-values-from-text-file-to-find-average-of-the-group-of-numbers#answer_784829

在 MATLAB Online 中打开

Try this —

T1 = readtable('https://www.mathworks.com/matlabcentral/answers/uploaded_files/734899/Sujata%20Dhar%20%20text%20file%20-%202021%2009%2010.txt', 'VariableNamingRule','preserve');
sigma = regexp(T1{:,6}, '\d*.\d*[^[cm]]','match');
T2 = table(T1{:,2}, str2double([sigma{:}]).', 'VariableNames',{'#','sigma'})
T2 = 36×2 table
#sigma_________

  0.62888
  0.91947
   1.6493
    2.285
   2.9204
   3.5172
   4.0523
   4.5663
   5.0203
   5.4476
   5.8295
   6.1764
   6.4962
   6.7936
   7.0645
     7.31
sigma_25to30 = T2.sigma(25:30)
sigma_25to30 = 6×1
7529
5772
8834
5924
2609
9016

or possibly —

sigma_25to30 = T2.sigma(T1.('#')>=25 & T2.('#')<=30)
sigma_25to30 = 6×1
5772
8834
5924
2609
9016
4905

I have no idea if this will work with the original text file because it never actually appeared as an uploaded/attached file. If there is a problem with it, use the one supplied here instead, since it works with my code.

Experiment to get different results.

.

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Star Strider 2021-9-10

在 MATLAB Online 中打开

My pleasure. It would have been helpful to have the file originally.

Try this —

T1 = readtable('https://www.mathworks.com/matlabcentral/answers/uploaded_files/734959/r1999_simulator.txt', 'VariableNamingRule','preserve')
T1 = 1375×7 table
Var1#Typememberreference epochsigma [unit]repeatability [unit]____________________________________________________________________________________

    NaN      0    {'PWL_CLK'}    {'AGGO'}    {'2021.05.18 17:00:00'}    {'0.62888 [cm]'}     {'1.99957 [cm]' }  
    NaN      1    {'PWL_CLK'}    {'AGGO'}    {'2021.05.18 18:00:00'}    {'0.91947 [cm]'}     {'4.93553 [cm]' }  
    NaN      2    {'PWL_CLK'}    {'AGGO'}    {'2021.05.18 19:00:00'}    {'1.64927 [cm]'}     {'8.35032 [cm]' }  
    NaN      3    {'PWL_CLK'}    {'AGGO'}    {'2021.05.18 20:00:00'}    {'2.28501 [cm]'}     {'10.70383 [cm]'}  
    NaN      4    {'PWL_CLK'}    {'AGGO'}    {'2021.05.18 21:00:00'}    {'2.92039 [cm]'}     {'12.11574 [cm]'}  
    NaN      5    {'PWL_CLK'}    {'AGGO'}    {'2021.05.18 22:00:00'}    {'3.51718 [cm]'}     {'12.86139 [cm]'}  
    NaN      6    {'PWL_CLK'}    {'AGGO'}    {'2021.05.18 23:00:00'}    {'4.05227 [cm]'}     {'12.85490 [cm]'}  
    NaN      7    {'PWL_CLK'}    {'AGGO'}    {'2021.05.19 00:00:00'}    {'4.56631 [cm]'}     {'12.34585 [cm]'}  
    NaN      8    {'PWL_CLK'}    {'AGGO'}    {'2021.05.19 01:00:00'}    {'5.02033 [cm]'}     {'11.29844 [cm]'}  
    NaN      9    {'PWL_CLK'}    {'AGGO'}    {'2021.05.19 02:00:00'}    {'5.44759 [cm]'}     {'10.22204 [cm]'}  
    NaN     10    {'PWL_CLK'}    {'AGGO'}    {'2021.05.19 03:00:00'}    {'5.82947 [cm]'}     {'9.16751 [cm]' }  
    NaN     11    {'PWL_CLK'}    {'AGGO'}    {'2021.05.19 04:00:00'}    {'6.17638 [cm]'}     {'8.30392 [cm]' }  
    NaN     12    {'PWL_CLK'}    {'AGGO'}    {'2021.05.19 05:00:00'}    {'6.49624 [cm]'}     {'7.89360 [cm]' }  
    NaN     13    {'PWL_CLK'}    {'AGGO'}    {'2021.05.19 06:00:00'}    {'6.79364 [cm]'}     {'7.92088 [cm]' }  
    NaN     14    {'PWL_CLK'}    {'AGGO'}    {'2021.05.19 07:00:00'}    {'7.06446 [cm]'}     {'8.59745 [cm]' }  
    NaN     15    {'PWL_CLK'}    {'AGGO'}    {'2021.05.19 08:00:00'}    {'7.30998 [cm]'}     {'9.59593 [cm]' }  
sigma = regexp(T1{:,6}, '\d*.\d*[^[cm]|[mas]|[cm/day]|[cm/day^2]]','match');
T2 = table(T1{1:end-1,2}, str2double([sigma{:}]).', 'VariableNames',{'#','sigma'})
T2 = 1374×2 table
#sigma_________

     0    0.62888
     1    0.91947
     2     1.6493
     3      2.285
     4     2.9204
     5     3.5172
     6     4.0523
     7     4.5663
     8     5.0203
     9     5.4476
    10     5.8295
    11     6.1764
    12     6.4962
    13     6.7936
    14     7.0645
    15       7.31
sigma_25to30 = T2.sigma(25:30)
sigma_25to30 = 6×1
    9.7529
    0.5772
    0.8834
    1.5924
    2.2609
    2.9016
mean_25to30 = mean(sigma_25to30)
mean_25to30 = 2.9947

This turned out to be a bit more of a challenge than I originally envisioned. (The last element of ‘sigma’ is empty.)

Experiment to get dfifferent resullts.

.

Sujata Dhar 2021-9-11

Thank you very much. :) :)

Star Strider 2021-9-11

As always, myu pleasure!

.

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