![]() ![]() Now, the whole construction is not very nice, I'm sure I could have used the commands in a more efficient way, so feel free to point that out. I have to use First because some of the output comes in angular brackets. half looks for the index of the point in the y data that is closest to half the minimum, and then FWHM calculates the difference between the x coordinate belonging to half the minimum and the actual minimum, and then multiplies by two. It is not all that elegant, but it works. The starting points were 'guessed' with the assistance of manipulate. In the following data was imported using ImportString from pastebin link.Ī non-linear model was developed. You could also search around the midpoint. I don't need a 100% perfect estimation this should be good enough. My plan was to find this position of the half-max point, find the corresponding xx, subtract the xx element of the position of the max, and multiply by two. ![]() My question is therefore if anyone seems a workaround, or an all around better idea. I've tried thinking of a solution to this, but it doesn't seem obvious to me. I'm not sure why exactly this is, but maybe there is some rounding going on, or incompatible outputs. ![]() Ironically, using Position)*0.5,1]] seems to come up empty. But I'm having some trouble finding the actual position of this point in the data. The first thing that came to mind was using Nearest)*0.5,1] which indeed returns a value of yyclosest to half of the FWHM. To find the full width half max, it is then probably easiest to look for a point closest to 1-(1-Min)*0.5 in yy. Say we call the data yy, then by using Min we've found the minimum. Now, this set has a minimum, which can be found rather easily. A picture of the data is given below, and the dataset itself can be found here (pastebin). I have a dataset of which I'm attempting to determine the full width at half maximum. ![]()
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