I am working on Google Tensorboard, and I'm feeling confused about the meaning of Histogram Plot. I read the tutorial, but it seems unclear to me. I really appreciate if anyone could help me figure out the meaning of each axis for Tensorboard Histogram Plot.
Sample histogram from TensorBoard
I came across this question earlier, while also seeking information on how to interpret the histogram plots in TensorBoard. For me, the answer came from experiments of plotting known distributions.
So, the conventional normal distribution with mean = 0 and sigma = 1 can be produced in TensorFlow with the following code:
import tensorflow as tf
cwd = "test_logs"
W1 = tf.Variable(tf.random_normal([200, 10], stddev=1.0))
W2 = tf.Variable(tf.random_normal([200, 10], stddev=0.13))
w1_hist = tf.summary.histogram("weights-stdev_1.0", W1)
w2_hist = tf.summary.histogram("weights-stdev_0.13", W2)
summary_op = tf.summary.merge_all()
init = tf.initialize_all_variables()
sess = tf.Session()
writer = tf.summary.FileWriter(cwd, session.graph)
sess.run(init)
for i in range(2):
writer.add_summary(sess.run(summary_op),i)
writer.flush()
writer.close()
sess.close()
Here is what the result looks like:
.
The horizontal axis represents time steps.
The plot is a contour plot and has contour lines at the vertical axis values of -1.5, -1.0, -0.5, 0.0, 0.5, 1.0, and 1.5.
Since the plot represents a normal distribution with mean = 0 and sigma = 1 (and remember that sigma means standard deviation), the contour line at 0 represents the mean value of the samples.
The area between the contour lines at -0.5 and +0.5 represent the area under a normal distribution curve captured within +/- 0.5 standard deviations from the mean, suggesting that it is 38.3% of the sampling.
The area between the contour lines at -1.0 and +1.0 represent the area under a normal distribution curve captured within +/- 1.0 standard deviations from the mean, suggesting that it is 68.3% of the sampling.
The area between the contour lines at -1.5 and +1-.5 represent the area under a normal distribution curve captured within +/- 1.5 standard deviations from the mean, suggesting that it is 86.6% of the sampling.
The palest region extends a little beyond +/- 4.0 standard deviations from the mean, and only about 60 per 1,000,000 samples will be outside of this range.
While Wikipedia has a very thorough explanation, you can get the most relevant nuggets here.
Actual histogram plots will show several things. The plot regions will grow and shrink in vertical width as the variation of the monitored values increases or decreases. The plots may also shift up or down as the mean of the monitored values increases or decreases.
(You may have noted that the code actually produces a second histogram with a standard deviation of 0.13. I did this to clear up any confusion between the plot contour lines and the vertical axis tick marks.)
#marc_alain, you're a star for making such a simple script for TB, which are hard to find.
To add to what he said the histograms showing 1,2,3 sigma of the distribution of weights. which is equivalent to the 68th,95th, and 98th percentiles. So think if you're model has 784 weights, the histogram shows how the values of those weights change with training.
These histograms are probably not that interesting for shallow models, you could imagine that with deep networks, weights in high layers might take a while to grow because of the logistic function being saturated. Of course I'm just mindlessly parroting this paper by Glorot and Bengio, in which they study the weights distribution through training and show how the logistic function is saturated for the higher layers for quite a while.
When plotting histograms, we put the bin limits on the x-axis and the count on the y-axis. However, the whole point of histogram is to show how a tensor changes over times. Hence, as you may have already guessed, the depth axis (z-axis) containing the numbers 100 and 300, shows the epoch numbers.
The default histogram mode is Offset mode. Here the histogram for each epoch is offset in the z-axis by a certain value (to fit all epochs in the graph). This is like seeing all histograms places one after the other, from one corner of the ceiling of the room (from the mid point of the front ceiling edge to be precise).
In the Overlay mode, the z-axis is collapsed, and the histograms become transparent, so you can move and hover over to highlight the one corresponding to a particular epoch. This is more like the front view of the Offset mode, with only outlines of histograms.
As explained in the documentation here:
tf.summary.histogram
takes an arbitrarily sized and shaped Tensor, and compresses it into a
histogram data structure consisting of many bins with widths and
counts. For example, let's say we want to organize the numbers [0.5,
1.1, 1.3, 2.2, 2.9, 2.99] into bins. We could make three bins:
a bin containing everything from 0 to 1 (it would contain one element, 0.5),
a bin containing everything from 1-2 (it would contain two elements, 1.1 and 1.3),
a bin containing everything from 2-3 (it would contain three elements: 2.2, 2.9 and 2.99).
TensorFlow uses a similar approach to create bins, but unlike in our
example, it doesn't create integer bins. For large, sparse datasets,
that might result in many thousands of bins. Instead, the bins are
exponentially distributed, with many bins close to 0 and comparatively
few bins for very large numbers. However, visualizing
exponentially-distributed bins is tricky; if height is used to encode
count, then wider bins take more space, even if they have the same
number of elements. Conversely, encoding count in the area makes
height comparisons impossible. Instead, the histograms resample the
data into uniform bins. This can lead to unfortunate artifacts in
some cases.
Please read the documentation further to get the full knowledge of plots displayed in the histogram tab.
Roufan,
The histogram plot allows you to plot variables from your graph.
w1 = tf.Variable(tf.zeros([1]),name="a",trainable=True)
tf.histogram_summary("firstLayerWeight",w1)
For the example above the vertical axis would have the units of my w1 variable. The horizontal axis would have units of the step which I think is captured here:
summary_str = sess.run(summary_op, feed_dict=feed_dict)
summary_writer.add_summary(summary_str, **step**)
It may be useful to see this on how to make summaries for the tensorboard.
Don
Each line on the chart represents a percentile in the distribution over the data: for example, the bottom line shows how the minimum value has changed over time, and the line in the middle shows how the median has changed. Reading from top to bottom, the lines have the following meaning: [maximum, 93%, 84%, 69%, 50%, 31%, 16%, 7%, minimum]
These percentiles can also be viewed as standard deviation boundaries on a normal distribution: [maximum, μ+1.5σ, μ+σ, μ+0.5σ, μ, μ-0.5σ, μ-σ, μ-1.5σ, minimum] so that the colored regions, read from inside to outside, have widths [σ, 2σ, 3σ] respectively.
I am implementing Lowe's method, "SIFT", for finding and describing features in an image.
I have found interest points, and now I have to describe them: Using Lowe's method, I have calculated the magnitude and gradient in an area around the keypoint, and created a Gauss weighted histogram, with 36 bins, each corresponding to an orientation of 10 degrees. For each keypoint, there is a histogram. Each bin is the sum of the weighted magnitude, in that direction. An example taken from aishack.in: http://www.aishack.in/static/img/tut/sift-orientation-histogram.jpg
Bins within 80% the size of the maximum bin, is made a new keypoint. After describing, it says in the paper: "Finally, a parabola is fit to the 3 histogram values closest to each peak to interpolate the peak position for better accuracy". I am not sure i get this.
In my understanding, it means the peak, the left, and the right value of that peak, will have a parabola fit, like this(be warned! Drawn free hand)
http://i.stack.imgur.com/7V8pb.jpg
and the orientation of the keypoint will be where the extremum of the parabola is. For instance: If the parabola fitted at 10-19, 20-29, and 30-39 (with 20-29 being the histogram peak), had extremum at a point, that reached in the 30-39, then this would be the orientation of that keypoint. Am i understanding this correctly? In this way, the orientation of the keypoint, can only be within 36 orientations
Another option: Same idea as above, only the histogram is no longer discrete: the extremum of the parapola will thus be a continuous value, and this value is assigned to the keypoint.
The idea of the parabola fitting is to find the peak with better than bin resolution. As you see in your example, the peak is at 20-29 (average 24.5) but the 10-19 bin is higher than the 30-39 bin. It's therefore likely that the precise peak should be below 24.5.
You can't have a non-discrete histogram, that defeats the point of a histogram. What you can have is overlapping bins: create a bin for 20-29, but also a bin for 21-30 and 22-31 etc. So the value 24 would map to 10 bins, from 15-24 to 24-35.
And when you increment a bin, you don't necessarily need to increment it by 1. You can also increment a bin by a variable amount, e.g. the distance from the given value to the edge of the bin. So 24 would add 1 to bin 16-25 but 4 to bin 20-29.
I'm newbie in cepstrum analysis. So that's the question.
I have signal with the length 4096 and sample rate 8000 Hz. I make FFT and get the array with the length 4096*2 (2*i position is for cosinus coeff, 2*i+1 position is for sinus coeff). Frequency step is (sampleRate/signalLength == 8000/4096). So, I can calculate frequency at i position this way: i*sampleRate/signalLength.
Then, I make the cepstrum transformation. I can't understand how to find quefrency step and how to find frequency for given quefrency.
The bin number of an FFT result is inversely proportional to the length of the period of a sinusoidal component in the time domain. The bin number of a quefrency result is also inversely proportional to the distance between partials in a series of overtones in the frequency domain (this distance often the same as a root or fundamental pitch). Thus quefrency bin number would be proportional to period or repeat lag (autocorrelation peak) of a harmonically rich periodic signal in the time domain.
I have a 48-bit (16 bits per pixel) image I've loaded with FreeImage. I'm trying to generate a histogram from this image without having to convert it to a 24-bit image.
This is how I understand histograms are calculated..
for (pixel in pixels)
{
red_histo[pixel.red]++;
}
Where pixel.red can be between 0 and 255. So there is a range from 0 to 255 on my histogram. But if there is 16 bits per pixel, it could be between 0 and 65535, which is too large to be displayed on a histogram.
Is there a standard way to calculate histograms with 48-bit (or higher) images?
You have to decide how many bins you need in the histogram. For eg. the Matlab histogram function takes these forms
imhist(I)
imhist(I, n)
imhist(X, map)
In the first case, the number of bins is by default used as 256. So, if you have 16bit input, these will be scaled down to 8 bit and split into 256 bin histogram.
In the second one, you can specify number of bins 'n'. Lets say you specify n=2 for your 16 bit data. Then, this will essentially split the histogram as [0-2^15, 2^15-2^16-1].
The third case is where you specify the map for each bin. ie you have to specify the ranges of the pixel values for each bin.
http://www.mathworks.com/help/images/ref/imhist.html
How you want to choose the number of bins depends on your requirement.
This Stack Overflow Question May have the answer you are looking for.
I do not know if there is a "standard" way.
If this is for display purposes you can scale back the pixels to keep the range from 0-255 for instance:
double scalingFactor = 255/65535;
for (pixel in pixels)
{
red_histo[(int)(scalingFactor * pixel.red)]++;
}
This will allow the upper range of the 16 bit pixel to come in at 255 and lower range of the 16 bit pixel to come in at 0.
I've got a 4096 samples long 44.1 kHz audio-clip. After applying the FFT to it I get 4096 frequency bands.
Each band would then span 10.77 Hz (44100 / 4096).
I've been told the 2nd half of the frequencies is conjugate symmetric to the first half.
Considering this is my calculation above correct or did I miss something important?
That's pretty much correct - for most common complex-to-complex FFTs with purely real inputs (i.e. all imaginary parts zero) the first N/2 output bins (0..2047 in your case) are typically the only bins that you will be interested in. The first bin is DC (0 Hz), and bin N/2 corresponds to Nyquist (Fs/2 = 22.05 kHz), which is not normally of interest. Bins above N/2 are just complex conjugate "mirror images" of the bottom N/2-1 bins.
See this answer for more details.