如何在python(numpy/scipy)中的嘈杂数据集中找到峰的FWHM? [英] How can I find the FWHM of a peak in a noisy data set in python (numpy/scipy)?

问题描述

我正在分析两条交叉线(如+号)的图像，并提取与其中一条线垂直的像素线(nx1 numpy数组).这给了我一个可以绘制的浮点值数组(代表颜色).我正在使用matplotlib绘制数据，得到了180至200之间的一堆嘈杂数据，中间有一个明显的峰值，峰值下降到100左右.

I am analyzing an image of two crossing lines (like a + sign) and I am extracting a line of pixels (an nx1 numpy array) perpendicular to one of the lines. This gives me an array of floating point values (representing colors) that I can then plot. I am plotting the data with matplotlib and I get a bunch of noisy data between 180 and 200 with a distinct peak in the middle that spikes down to around 100.

我需要找到此数据的FWHM.我认为我需要首先过滤噪声，所以我使用了高斯滤波器，可以平滑我的数据，但是它的顶部仍然不是超级平坦.

I need to find FWHM of this data. I figured I needed to filter the noise first, so I used a gaussian filter, which smoothed out my data, but its still not super flat at the top.

1. 我想知道是否有更好的方法来过滤数据.

1. I was wondering if there is a better way to filter the data.

如何找到此数据的FWHM?

How can I find the FWHM of this data?

如果可能的话，我只想使用numpy，scipy和matplotlib.

I would like to only use numpy, scipy, and matplotlib if possible.

以下是原始数据:

以下是过滤后的数据:

Here is the filtered data:

推荐答案

我最终没有使用任何过滤器，而是使用了原始数据. 我使用的过程是:

I ended up not using any filter, but rather used the original data. The procedure I used was:

• 找到最小和最大点并计算difference = max(arr_y) - min(arr_y)
• 找到最大半数(在我的情况下是最小半个数)HM = difference / 2
• 找到了距HM最近的数据点:nearest = (np.abs(arr_y - HM)).argmin()
• 计算了最接近距离和最小值之间的距离(这给了我HWHM)
• 然后简单地乘以2即可得到FWHM

• Found the minimum and maximum points and calculated difference = max(arr_y) - min(arr_y)
• Found the half max (in my case it is half min) HM = difference / 2
• Found the nearest data point to HM: nearest = (np.abs(arr_y - HM)).argmin()
• Calculated the distance between nearest and min (this gives me the HWHM)
• Then simply multiplied by 2 to get the FWHM

我不知道(或认为)这是最好的方法，但是它可以工作，并且根据比较似乎相当准确.

I don't know (or think) this is the best way, but it works and seems to be fairly accurate based on comparison.

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