From char to timetable

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Adriano
Adriano am 26 Feb. 2019
Beantwortet: Adriano am 26 Feb. 2019
Hi everybody! I have a char variable of 1x446405. The structure of the variable is always the same like below:
2019-02-25 0.870134 -0.228498 -0.058666 -0.386660 0.083051 0.209336 0.327385 0.440973 0.551807 0.660402 0.766645
2019-02-24
2019-02-23
2019-02-22 0.870663 -0.221149 -0.048423 -0.384026 0.093777 0.219200 0.335723 0.447541 0.556605 0.663568 0.768380
2019-02-21 0.909650 -0.218755 -0.041825 -0.383403 0.105566 0.236380 0.358028 0.474450 0.587488 0.697766 0.805262
2019-02-20 0.874319 -0.220516 -0.047655 -0.383219 0.094916 0.220812 0.337824 0.450094 0.559548 0.666825 0.771874
2019-02-19 0.885250 -0.236768 -0.069948 -0.388208 0.072871 0.202418 0.324662 0.442630 0.557599 0.669865 0.779220
2019-02-18 0.877976 -0.250780 -0.088779 -0.394880 0.052791 0.183115 0.307173 0.427380 0.544648 0.659067 0.770328
2019-02-17
2019-02-16
2019-02-15 0.897203 -0.220365 -0.051245 -0.375429 0.092066 0.221087 0.342288 0.459007 0.572706 0.683782 0.792080
2019-02-14 0.864219 -0.224309 -0.052387 -0.386413 0.089209 0.214173 0.330341 0.441876 0.550706 0.657468 0.762102
2019-02-13 0.898676 -0.233793 -0.063130 -0.389390 0.082196 0.213365 0.336644 0.455262 0.570627 0.683122 0.792598
2019-02-12 0.913564 -0.205047 -0.021548 -0.379842 0.127451 0.257085 0.376145 0.489421 0.599253 0.706544 0.811406
2019-02-11 0.914752 -0.217810 -0.039717 -0.383609 0.108531 0.239985 0.362113 0.478895 0.592196 0.702663 0.810286
2019-02-10
I would like to get a timetable in which the rownames are the dates and the columns are the values. I would like also to eliminate the dates without values like 2019-02-24, 2019-02-23, 2019-02-17, 2019-02-16 and 2019-02-10. Many thanks!

Akzeptierte Antwort

Sean de Wolski
Sean de Wolski am 26 Feb. 2019
How are you creating the char? Reading it from a file? If you're reading it from a file, then look at importing with fixed width. Attach the file for better help.
Otherwise, something along the lines of the following.
lines = splitlines(string(s))
Which should give you separate strings for each line. Then any line with
empty = strlength(line)<12
Has no data so remove them
lines = lines(~empty)
Now create the timetable by parsing the strings (i.e. create a datetime from the first few characters, and split the rest on white space and convert to double. Something along the lines of the following.
dates = datetime(extractBefore(lines,12))
values = double(split(extractAfter(lines,12)))
timetable(dates,values)

Weitere Antworten (1)

Adriano
Adriano am 26 Feb. 2019
Simply great!!! The solution is what i wanted. I didn't know the splitlines function and it's very useful. Many thanks!!

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