## Histogram (automatic binning)

There are different methods for choosing widths of bins for a histogram, like Freedman-Diaconis rule. In Mathematica, without choosing a specific method, with the command ~Histogram[data]~, we can produce a histogram for a set of data. My question is: How does Mathematica decide which method is better for a given data?

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## Plotting a normal curve over a histogram

I am teaching a basic introduction to normal curves, and the textbook introduces the topic with the idea of rolling five dice and recording the sum. Do this many times and create a histogram of the results. (which looks more and more like the normal curve as the number of trials increases) My "code" to simulate this for 10000 rolls is below.

``rolls = Table[Table[RandomChoice[Range], {5}], {10000}]; sums = Total /@ rolls; Histogram[sums, {0.5, 31.5, 1}] ``

I am way out of my depth here, but wondered if anyone could help with how to plot the "perfect" normal curve that would match the histogram?

I have the following to plot a normal curve, but for it to match the dice example, I need the mean and standard deviation, and how no idea what that would be.

``Plot[PDF[NormalDistribution[0, 1], x], {x, -3, 3}, Ticks -> None,   Axes -> None] ``

Any help is appreciated.

## One Histogram or three in “hspec” option

I have a data set that naturally breaks into three subsets that I want to display as stacked histograms. If I use an hspec option like "Probability", it seems apply to each separately. I would prefer to see it applied to the original, union of the three. Is this possible?

## Recording a histogram in a tree exhibits strange best case

The task is to record a histogram from a streaming data source.

One data point is, say, a 16 bit integer. The maximum multiple of one data point before the stream ends, is < 2^32. The main restriction is that there isn’t enough memory to have an array of counters which completely covers the range of the data points (65536 counters of 0-2^32), much less record all single data points and bulk-process them afterwards. In the case where the dispersion of the data source covers the complete possible 16-bit range, the result is not of interest, i.e. this marks a histogram type which is not processed further. Valid histograms consist of only a few clusters where the majority of points are found. Data is roughly continuous which has a very valuable consequence: new outliers can, without loss of detection capability) supersede old outliers (i.e. those with the lowest count) so that new clusters can enter the histogram even when the available memory is full. This is very rare however, the data source is in a way nice to track; just the number and location of the clusters is unpredictable. The following is a try of mine with a binary tree recording the multitude of each data point (I didn’t implement the substitution, just the insert). The special property of the tree is that it tries to move points with a high occurrence number close to the root, in hope for a short search path in the majority cases. The boolean predicate is the function `exchange_p` which compares the weight of a root to its left or right child. In case the predicate is true, an exchange is executed, possibly invoking further exchanges lateron.

The idea is that it is favorable to not exchange tree nodes at every insert but to let the node count run up to a certain multiple of its parent before an exchange is initiated, thereby avoiding unnecessary oscillations of nodes. Things worked out very differently, tho.

The problem exhibits a strange behaviour, with `QUOTA` around 0.8 (i.e. exchange if root < 0.8*child) for the overall minimum number of operations. Is there some theory which amalgates O-notation with stochastics for special input distributions?

You can run the below example (it has real world data in it), the file is completely self-contained and should run out of the box.

``#include <cstdio> #include <cstdint> #include <cstdlib>   int exchg; int ins; int tree_ins;  float QUOTA = 1.0f;  void error_func (const char *s) {   printf ("%s", s);   exit (1); }  typedef enum { LEFT_CHILD, RIGHT_CHILD, NO_CHILD } child_side_t;    typedef int16_t element_t; typedef uint32_t counting_t;   struct bin_tree_counting_node_t {   bin_tree_counting_node_t *left, *right;   counting_t cnt;   element_t element; };   void print_node (bin_tree_counting_node_t* root) {   printf ("%i\t%u\n", root->element, root->cnt); }  void print_tree (bin_tree_counting_node_t* root, int lvl) {   if (root == NULL)     return;   print_tree (root->left, lvl+1);   // printf ("Level %d: ", lvl);   print_node (root);   print_tree (root->right, lvl+1); }     bool exchange_p (bin_tree_counting_node_t* root,         bin_tree_counting_node_t* child,         child_side_t side) {   return (root->cnt < QUOTA * child->cnt); }  bin_tree_counting_node_t* tree_insert_tree (bin_tree_counting_node_t* tree_a,          bin_tree_counting_node_t* tree_b) {   tree_ins++;      if (tree_a == NULL)     return tree_b;   else if (tree_b == NULL)     return tree_a;   else if (exchange_p (tree_a, tree_b, NO_CHILD)) {     if (tree_a->element < tree_b->element)        tree_b->left = tree_insert_tree (tree_a, tree_b->left);     else       tree_b->right = tree_insert_tree (tree_a, tree_b->right);     return tree_b;   }   // Case for a > b coincides with a ~ b   // else if (exchange_p (tree_b, tree_a, NO_CHILD)) {   //   if (tree_a->element < tree_b->element)    //     tree_a->right = tree_insert_tree (tree_a->right, tree_b);   //   else   //     tree_a->left = tree_insert_tree (tree_a->left, tree_b);   //   return tree_a;   // }   else {     if (tree_a->element < tree_b->element)        tree_a->right = tree_insert_tree (tree_a->right, tree_b);     else       tree_a->left = tree_insert_tree (tree_a->left, tree_b);     return tree_a;   }       }  bin_tree_counting_node_t* exchange (bin_tree_counting_node_t* root,      child_side_t side) {   exchg++;      // Exchange root with its child   bin_tree_counting_node_t* child;   if (side == LEFT_CHILD) {     child = root->left;     root->left = NULL;     child->right = tree_insert_tree (root, child->right);   }   else {     child = root->right;     root->right = NULL;     child->left = tree_insert_tree (root, child->left);   }   return child; }  bin_tree_counting_node_t* tree_insert (bin_tree_counting_node_t* root,         element_t elem) {   ins++;      if (root == NULL) {     bin_tree_counting_node_t* new_node = (bin_tree_counting_node_t*)malloc (sizeof (bin_tree_counting_node_t));     if (new_node == NULL) { error_func ("Memory exhausted!"); }     new_node->element = elem;     new_node->cnt = 1;     new_node->right = new_node->left = NULL;     return new_node;   }      if (elem == root->element) {     root->cnt++;     if (root->cnt == 0) { error_func ("Counting overflow! Very large amount of inserts >2^32!"); }   }   else if (elem < root->element) {     root->left = tree_insert (root->left, elem);     if (exchange_p (root, root->left, LEFT_CHILD))       return exchange (root, LEFT_CHILD);   }   else {     root->right = tree_insert (root->right, elem);     if (exchange_p (root, root->right, RIGHT_CHILD))       return exchange (root, RIGHT_CHILD);   }   return root; }  void free_tree(bin_tree_counting_node_t* root) {   if (root == NULL)     return;   free_tree (root->left);   free_tree (root->right);   free (root); }  int main (void) {   element_t sample[] =     {       17060,17076,17076,17060,17092,17028,17076,17060,17060,17076,17060,17076,17060,17060,17140,17140,17124,17124,       17124,17140,17108,17140,17124,17124,17108,17108,17124,17124,17140,17092,17124,17124,17140,17140,17124,17156,       17140,17108,17156,17140,17124,17124,17124,17108,17124,17124,17124,17092,17140,17092,       17124,17108,17156,17124,17156,17140,17124,17172,17124,17108,17124,17108,17108,17108,17108,17124,17108,17124,       17108,17108,17124,17124,17140,17124,17108,17092,17108,17108,17092,17124,17108,17108,17124,17108,17124,17108,       17124,17140,17156,17124,17108,17108,17124,17124,17124,17140,17092,17140,17124,       17108,17124,17124,17124,17124,17108,17124,17108,17124,17108,17108,17108,17124,17108,17124,17124,17108,17108,       17124,17108,17124,17140,17124,17124,17108,17108,17140,17140,17124,17108,17140,17124,16291,16339,16339,16307,       16323,16259,16275,16275,16259,16388,16388,16355,15795,15731,16195,16179,16179,       15715,14467,14643,14851,17284,17012,16147,15155,15203,16131,15331,14691,14739,14739,14755,14723,14739,14707,       14771,14739,14707,14691,14531,14787,14563,15587,15907,15907,15923,15907,13123,13107,13091,13267,13091,13187,       13091,14643,15875,15907,16964,16404,16227,15219,14771,14771,14803,14787,14803,       14787,14803,14803,14755,14755,14771,14755,14723,14723,14739,14739,14963,16884,16868,15827,13075,13123,13091,       12979,13043,13219,13075,13059,13075,13123,15779,15875,16916,17028,15155,14675,14707,14691,14707,14691,14691,       14691,14707,14691,14691,14691,14675,14707,14691,14675,14595,14595,14595,14563,       14547,14579,14611,14547,14515,14611,14595,14611,14595,14659,14659,14627,14643,14643,14659,14643,14643,14643,       14659,14659,14643,14643,14611,14659,14611,14659,14659,14659,14643,14643,14643,14643,14611,14643,14627,14675,       14627,14643,14531,14531,14499,15187,16884,16932,15891,15923,13107,13091,13043,       13059,13075,13027,13027,13059,13059,13587,13059,15763,16900,16932,14899,14595,14627,14659,14691,14691,14691,       14691,14659,14675,14675,14675,14675,14691,14675,14691,14675,14611,14643,15107,16788,16964,14835,13363,13059,       13075,13075,13043,13043,13043,13075,13059,13059,15475,15715,16884,16932,16964,       14707,14643,14675,14643,14611,14611,14611,14611,14627,14611,14611,14627,14627,14611,14595,14595,14595,14563,       15027,16756,16932,14595,13059,13075,13059,13059,13011,13027,13059,13059,13043,13059,15683,16852,16948,16067,       15267,14771,15875,16964,16996,16980,17012,17012,17028,17028,17076,17060,17044,       17060,17076,17076,17092,17076,17092,17092,17108,17124,17092,17108,17124,17108,17124,17124,17108,17140,17124,       17108,17124,17108,17028,17124,17124,17092,17124,17108,17124,17108,17124,17092,17108,17108,17108,17124,17124,       17140,17108,17124,17108,17124,17108,17108,17156,17124,17108,17092,17108,17092,       17108,17108,17124,17124,17108,17108,17108,17124,17108,17124,17108,17124,17124,17108,17124,17124,17124,17124,       17124,17092,17140,17108,17124,17124,17140,17028,17124,17028,17108,17108,17124,17124,17124,17124,17108,17108,       17124,17140,17044,17108,17028,17124,17124,17124,17108,17124,17124,17108,17124,       17092,17124,17124,17124,17108,17124,17124,17140,17124,17140,17140,17140,17124,17044,17124,17092,17140,17124,       17140,17124,17124,17124,17108,17124,17124,17124,17124,17124,17140,17076,17140,17140,17108,17140,17124,17140,       17140,17108,17108,17124,17108,17124,17124,17124,17108,17140,17124,17140,17060,       17124,17124,17188,17108,17124,17124,17140,     };    for (QUOTA = 0.1; QUOTA < 3; QUOTA += 0.1) {     ins = tree_ins = exchg = 0;          bin_tree_counting_node_t *tree = NULL;      for (int i=0; i<sizeof sample/sizeof sample; i++) {       tree = tree_insert (tree, sample[i]);     }     printf ("Q:\t%f\tins:\t%d\t tree_ins:\t%d\t exchg:\t%d\t sum:\t%d\n", QUOTA, ins, tree_ins, exchg, ins + tree_ins + exchg);     //print_tree (tree, 0);     //printf ("\n");     free_tree (tree);   }   return 0; } ``
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## Complexity of find histogram bins vs convex hull

For a list of n 2d points, finding the convex hull vertex takes O(n log(n)) time. And O(n) time if it’s sorted lexicon order.

Meanwhile What’s the complexity of finding the histogram bin edges of k bins on both axis ?

Which one is faster ?

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## Histogram Memory Allocation Failure

I have a list of 3256 pairs of data with which I plotted a scatter plot, no problem. But when I tried to take the ratio with the two elements of each pair and do a Histogram of the ratios, I got a Memory Allocation Failure. Please help.

ListPlot[ABC]

Histogram[MapThread[ If[#2 == 0 || #1 == 0, 0, #1/#2] &, {ABC[[All, 1]], ABC[[All, 2]]}]

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## Number of bins in histogram after apply CLAHE from cv2 in python

I try to use CLAHE in cv2 to process some of my experiment images. I thought if I set the tile size same as the size of image, it will just do normal histogram equalization. However, it turns out if I increase the tile size, the number of bins in histogram of output will decrease. When I set the tile size 60*60 (same as my image), there are only 4 kinds of pixel values in the output (no matter how I set the contrast limit). This leads to loss of details in the image. Is there any other method I can do histogram equalization while maintaining the number of bins? Thank you very much!!!

The number of bins decreases with the increase of the tile size tried different values for contrast limit but didn’t work

``import cv2  cv2.imread('1.png') clahe = cv2.createCLAHE(clipLimit=limit, tileGridSize=tilesize) img_new = clahe.apply(img)  plt.figure() plt.hist(img_new.ravel(),256,[0,256]) ``

## This code belongs to Equalization with Histogram

My Code brings up this error that “module ‘scipy.misc’ has no attribute ‘fromimage’ “. I have been working on this installed all the packages, still I am not getting any positive result.

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## Highlight individual datapoints in histogram bin

I have a graph showing the load time distribution for websites within an industry. The x-axis is showing the load time buckets (<1 s, 1-2 s, 2-3 s, 4-5 s and so on) and the y-axis is showing number of sites. The idea is that a user should be able to get a holistic view of how websites within an industry performs. On this graph I would also like to display which load time bucket one or two specific sites belongs to so that the user can see how these sites compare to each other and the industry in general.

What I have so far is this…  However, this is somewhat of a misrepresentation of the data since a site is not an entire bucket but rather part of it. Another issue is that when multiple sites end up in the same bucket (which is very common) there is no easy way of coloring the bar.

Is there a better way of highlighting which bucket a single data point belongs to?

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## camera methods(library or function ) to plot histogram after calculation?

what methods(library or function ) cameras use to plot histogram after calculation of necessary data ? I was thinking of using gnuplot or koolplot , but these methods seems little slow as I have to plot histogram in real time for videos .(what other problems possibly I can face using these tools ?)

I am not asking for any algorithm or code , just want to know which tool cameras use to plot histogram so fast .

Are there any tools like these , which will provide good speed + good visualisation to plot histogram ? what cameras really use ? please provide me any source or link, really stuck for a while now .

why downvote please explain me : am I not clear with my quetion ? : is there lack of information or some other problem?

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