Visual walkthrough — 2D plots — line, scatter, bar, histogram, contour, imshow

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Step 1 — What data even is (dots on a number line)

WHAT. We have a list of measured numbers. Call them $x_0,x_1,\dots ,x_{n-1}$. Here the tiny subscript is just an address: $x_0$ is the first number, $x_1$ the second, and $n$ is how many numbers we have in total.

WHY. Before we can bucket anything, we must see the raw values. A histogram's whole job is to summarise this cloud of dots — so we look at the cloud first, undecorated.

PICTURE. Below, eight sample heights (in cm) are drawn as dots on a single horizontal line. No bars yet — just where each number lives.

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Step 2 — Fencing off the territory: min, max, and the range

WHAT. We find the smallest value $\min$ and the largest $\max$, then look at the stretch between them: the range $R=\max -\min$.

WHY. A histogram needs a fixed playing field. Every bucket must live inside $[\min ,\max ]$, otherwise some numbers would have nowhere to go. The range $R$ is the total width we must divide up.

PICTURE. Two vertical fences drop at $\min$ and $\max$. The green double-arrow is $R$ — the distance the fences enclose.

$R=\underset{\text{right fence}}{\underbrace{\max }}-\underset{\text{left fence}}{\underbrace{\min }}$

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Step 3 — Choosing $B$ buckets and slicing the range

WHAT. We pick a number of buckets, $B$ (the bins). We cut the range into $B$ equal slices. Each slice has width $h=\frac{R}{B}=\frac{\max -\min }{B}.$

WHY divide? We chose equal-width buckets so that every bar is comparable — a tall bar means "more numbers here," never "wider bucket here." Division is the tool that answers "if I share this length $R$ fairly among $B$ buckets, how much does each get?" — exactly like splitting a chocolate bar into equal pieces.

PICTURE. Same fences as Step 2, now chopped by $B-1$ interior tick marks into $B=4$ equal slices, each labelled with its width $h$.

$\underset{\text{one bucket’s width}}{\underbrace{h}}=\frac{\stackrel{\text{total range }R}{\overbrace{\max -\min }}}{\underset{\text{how many buckets}}{\underbrace{B}}}$

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Step 4 — Which bucket does a number fall in?

WHAT. Bucket $j$ (counting $j=0,1,\dots ,B-1$) covers the interval $[\min +jh,\min +(j+1)h).$ The square bracket $[$ means "including this end"; the round bracket $)$ means "up to but not including this end." A number $x_i$ lands in bucket $j$ when $\min +jh\le x_i<\min +(j+1)h.$

WHY the half-open interval? Every number must belong to exactly one bucket — no number may sit on a fence and get counted twice or zero times. Making the left edge inclusive ($\le$) and the right edge exclusive ($<$) glues the buckets together with no overlap and no gaps.

PICTURE. One highlighted bucket $j$. Its left edge is a solid dot (included), its right edge a hollow dot (excluded). A test point $x_i$ is shown obeying $\text{left}\le x_i<\text{right}$.

$\underset{\text{bucket’s left edge (solid)}}{\underbrace{\min +jh}}\le x_i<\underset{\text{next bucket’s left edge (hollow)}}{\underbrace{\min +(j+1)h}}$

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Step 5 — Counting: the height $c_j$

WHAT. For each bucket $j$ we simply count how many data points landed inside it. That count is the bucket's height: $c_j=\#\{x_i:\min +jh\le x_i<\min +(j+1)h\}.$ The symbol $\#\{\dots \}$ means "the number of items in this set" — literally "count them."

WHY. This is the payoff. A histogram deliberately forgets the exact value of each point and remembers only how crowded each bucket is. Crowding is what reveals the shape of the data.

PICTURE. The eight dots drop straight down into their buckets and stack up. The stack height in bucket $j$ is $c_j$. This is the first time you see actual bars.

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Step 6 — From counts to a density (making the area equal 1)

WHAT. Raw counts depend on how many samples you took — 5000 draws give taller bars than 50. To compare shapes fairly (and to overlay a theoretical curve), we rescale each bar's height to ${\text{density height}}_j=\frac{c_j}{nh}.$

WHY this exact divisor $nh$? We want the total area of all bars to equal $1$ (the meaning of a probability density). The area of bar $j$ is height $\times$ width: ${\text{area}}_j=\underset{\text{new height}}{\underbrace{\frac{c_j}{nh}}}\times \underset{\text{width}}{\underbrace{h}}=\frac{c_j}{n}.$ Sum over all buckets: ${\sum }_j\frac{c_j}{n}=\frac{1}{n}{\sum }_j{c}_j=\frac{n}{n}=1.✓$ Dividing by $h$ cancels the width, and dividing by $n$ cancels the sample size — so the shape survives but the total area is pinned to 1.

PICTURE. Left panel: the count histogram (tall, height depends on $n$). Right panel: the same shape rescaled so its total area is 1, with a theoretical bell curve laid on top — now they live on the same vertical scale.

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Step 7 — Edge & degenerate cases (never leave the reader stranded)

WHAT / WHY / PICTURE — three things that break the naive formula:

1. All values identical ($\min =\max$). Then $R=0$ and $h=R/B=0$ — division gives a zero-width bucket, undefined. What tools do: Matplotlib nudges the fences apart (e.g. to $[\min -0.5,\max +0.5]$) so one lonely bar of height $n$ appears.
2. A value exactly on the far-right fence ($x_i=\max$). The half-open rule $<$ (Step 4) would drop it — so the last bucket is made a special closed interval $[\dots ,\max ]$ to catch it.
3. A value outside $[\min ,\max ]$ because you forced a range with range=(a,b). Points beyond the fences are simply not counted — the areas no longer sum to 1. Watch for silently missing data.

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The one-picture summary

Everything above compressed into one flow: dots → fences → equal slices → count → rescale to area 1.

Recall Feynman: retell the whole walkthrough in plain words

I start with a handful of numbers and just poke them onto a number line as dots (Step 1). I find the smallest and biggest and put fences there — the gap between fences is my "range" (Step 2). I decide how many buckets I want, and chop the range into that many equal-width slices; each slice's width is range-divided-by-number-of-buckets (Step 3). I glue the buckets edge-to-edge with a rule — a number belongs to a bucket if it's at-or-past the left edge but strictly before the right edge — so nobody is counted twice (Step 4). Then I let every dot fall into its bucket and I count the pile; the pile height is the bar (Step 5). Because tall piles just mean I took more samples, I shrink each bar by dividing by (number of samples × bucket width) so the total area becomes exactly 1 — now it's a probability density I can compare to a theory curve (Step 6). Finally I check the weird cases: all-equal numbers, a value sitting right on the last fence, and values pushed outside a forced range (Step 7). That's the whole histogram, no magic.

Recall

What does $h$ mean and how is it computed? ::: The width of one bin, $h=(\max -\min )/B$. Why divide the height by $nh$ for a density? ::: So that bar area = $c_j/n$ and all areas sum to 1. Why is the bin interval half-open $[,)$? ::: So every point lands in exactly one bin — no overlap, no gaps. What does $c_j$ count? ::: How many data points fall inside bin $j$.

Related: NumPy meshgrid and broadcasting · Colormaps and perceptual uniformity (viridis) · Matplotlib Figure vs Axes object model · Seaborn — statistical plotting wrappers