Worked examples — 2D plots — line, scatter, bar, histogram, contour, imshow

You have read the parent note and met the six plot types. Now we grind through every messy case you will actually hit: the degenerate ones (empty data, one bin, flat surface), the sign/orientation traps (upside-down images, negative extents), the limiting cases (many bins vs few), a real-world word problem, and an exam twist. Each example first asks you to guess, then shows the reasoning step-by-step, then verifies by plugging numbers back.

We will use only ideas already defined: a plot is a mapping from numbers to pixels; a histogram counts samples into bins of width $h=\frac{\max -\min }{B}$; density=True divides each count by $nh$ so total area is $1$; a contour at level $c$ is the set $\{(x,y):f(x,y)=c\}$; imshow shows an array as colored cells with Z[row, col] = Z[y, x].

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The scenario matrix

Every plotting bug or exam question lives in one of these cells. The examples below are labelled with the cell they cover.

#	Cell class	Concrete trap / question	Example
C1	Line — resolution limit	few points look jagged, many look smooth	Ex 1
C2	Histogram — bin arithmetic	which bin does a value land in? edge rule	Ex 2
C3	Histogram — normalization	does density=True really give area $=1$?	Ex 3
C4	Histogram — degenerate	$B=1$ bin, or all samples identical	Ex 4
C5	Bar — geometry	left edge, width, gaps, negative height	Ex 5
C6	Contour — the level set	which $(x,y)$ are on level $c$? radius formula	Ex 6
C7	imshow — orientation signs	origin upper vs lower, row↔y flip	Ex 7
C8	imshow — extent mapping	array index → real coordinate, incl. negatives	Ex 8
C9	Real-world word problem	temperature log: choose the right plot + numbers	Ex 9
C10	Exam twist	mix meshgrid shape + contour count	Ex 10

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Ex 1 — Line resolution (cell C1)

Forecast: Guess the number of segments and whether 5 points can look smooth.

1. Find the spacing. With $n=5$ points over $[0,2\pi ]$, the gap between consecutive points is $\Delta x=\frac{2\pi -0}{n-1}=\frac{2\pi }{4}=\frac{\pi }{2}.$ Why this step? np.linspace(a, b, n) puts points at the two ends and $n-2$ inside, so there are $n-1$ gaps — not $n$. Dividing by $n-1$ is the whole trick.
2. List the points. $x=\left(0,\frac{\pi }{2},\pi ,\frac{3\pi }{2},2\pi \right)$, giving $y=(0,1,0,-1,0)$. Why this step? These are exactly the peaks/zeros of sine, so the "curve" becomes a zig-zag of straight lines between them.
3. Count segments. A line plot draws ${\text{segment}}_i:(x_i,y_i)\to (x_{i+1},y_{i+1})$ for $i=0,\dots ,n-2$, i.e. $n-1=4$ segments. Why this step? The parent formula: $n$ points $\Rightarrow n-1$ connectors. Look at the red zig-zag in the figure — 4 straight pieces, no curve at all.

Verify: $\Delta x=\pi /2\approx 1.5708$. Segment count $=5-1=4$. The eye only sees a smooth sine once segments are tiny (Ex needs ~200 points, so ~199 segments). Units: $x$ in radians, $y$ dimensionless. ✔

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Ex 2 — Which bin? (cell C2)

Forecast: Guess whether $4.0$ lands in the same bin as $3.9$.

1. Bin width. $h=\frac{\max -\min }{B}=\frac{10-0}{5}=2$. Why this step? Everything depends on $h$; get it wrong and every index shifts.
2. Bin edges. $[0,2),[2,4),[4,6),[6,8),[8,10)$ — indices $j=0,1,2,3,4$. Why this step? Each interval is half-open (closed left, open right), so a value exactly on an edge goes to the bin on its right, except the very last edge.
3. Place the values. The formula for the index is $j=\lfloor \frac{x-\min }{h}\rfloor$.
   • $x=3.9$: $\lfloor 3.9/2\rfloor =\lfloor 1.95\rfloor =1$ → bin 1 $[2,4)$.
   • $x=4.0$: $\lfloor 4.0/2\rfloor =\lfloor 2.0\rfloor =2$ → bin 2 $[4,6)$.
   • $x=10.0$: naively $\lfloor 10/2\rfloor =5$, but there is no bin 5. Matplotlib puts the max value into the last bin, so $x=10.0$ → bin 4 $[8,10]$. Why this step? This is the classic "edge counted twice or lost" trap. The half-open rule stops double-counting; the last bin is closed on the right so the maximum isn't dropped.

Verify: $3.9$ and $4.0$ land in different bins (1 vs 2) — the boundary at $4$ belongs to the right bin. And $10.0$ is rescued into bin 4, not an illegal bin 5. ✔

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Ex 3 — Does density=True give area 1? (cell C3)

Forecast: Guess the total area before computing — it should be a clean number.

1. Bin width. $h=\frac{4-0}{4}=1$. Why this step? Area of a bar $=$ height $\times h$; we need $h$ to turn heights into areas.
2. Density heights. Each becomes $\frac{c_j}{nh}=\frac{c_j}{200\cdot 1}$: $\left(\frac{50}{200},\frac{90}{200},\frac{40}{200},\frac{20}{200}\right)=(0.25,0.45,0.20,0.10).$ Why this step? Dividing by $nh$ is the parent's normalization: it converts counts into a probability density so we can overlay a theoretical PDF (Probability density functions and normalization).
3. Sum the areas. Area $={\sum }_j({\text{height}}_j)\cdot h=(0.25+0.45+0.20+0.10)\cdot 1=1.00$. Why this step? Because ${\sum }_j\frac{c_j}{nh}\cdot h=\frac{1}{n}{\sum }_j{c}_j=\frac{200}{200}=1$. The counts must sum to $n$, so the areas must sum to 1.

Verify: Counts sum: $50+90+40+20=200=n$ ✔. Total area $=1.0$ ✔. Note heights are not probabilities themselves (bin 1 height $0.45>$ its probability $90/200=0.45$ only because $h=1$; with $h=0.5$ they'd differ). ✔

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Ex 4 — Degenerate histograms (cell C4)

Forecast: Guess whether $B=1$ makes the area still equal $1$, and whether identical samples crash.

1. (a) One bin. With $B=1$ the whole range $[\min ,\max ]$ is one bin: raw height $=n=200$. Why this step? Every sample lands in the only bin, so the count is all of them.
2. Density for one bin. $h=\max -\min$; height $=\frac{n}{nh}=\frac{1}{h}$, and area $=\frac{1}{h}\cdot h=1$ still. Why this step? Normalization is designed to give area 1 for any $B$ — even the useless $B=1$ that shows no shape.
3. (b) All identical. Then $\min =\max =7$, so $h=0$ — division by zero looms. NumPy's fix: it expands the range to $[6.5,7.5]$ (a unit-wide window around the value) so $h>0$ and the count $200$ goes into one bin without crashing. Why this step? A degenerate zero-width range would make every height infinite; the auto-padding keeps the plot finite and drawable.

Verify: (a) raw height $=200$; density area $=1$ regardless of $h$ ✔. (b) NumPy uses range $[6.5,7.5]$, so $h=1$ and the single count is $200$ ✔ — no exception thrown.

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Ex 5 — Bar geometry with a negative value (cell C5)

Forecast: Guess whether the negative bar points down and where its top/bottom sit.

1. Position. Category $k=1$ ("B") sits at integer $x=1$. Bar spans $[k-w/2,k+w/2]$ with $w=0.6$: left $=1-0.3=0.7$, right $=1+0.3=1.3$. Why this step? Parent anatomy: a bar is a rectangle centered on the category's integer slot.
2. Height sign. Height $=-3$ means the rectangle grows downward from the baseline $y=0$: its top is at $y=0$, its bottom at $y=-3$. Why this step? bar draws from the baseline (default 0) to the value; a negative value simply flips the direction below the axis.
3. Gap between bars. Neighbours are 1 unit apart (integers 0,1,2); each occupies width $0.6$, so the empty gap is $1-0.6=0.4$. Why this step? Width $<1$ is exactly what makes categories read as separate blocks — the parent's "WHY width matters".

Verify: Left $=0.7$, right $=1.3$ ✔. Bar B occupies $y\in [-3,0]$ ✔. Gap $=0.4>0$ so bars don't touch ✔.

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Ex 6 — The contour level set (cell C6)

Forecast: Guess the shape of the highest level's contour — a curve or a point?

1. Write the level equation. The contour at level $c$ is $\{(x,y):f(x,y)=c\}$, i.e. $e^{-(x^2+y^2)}=0.5$. Why this step? A contour is a set of points solving an equation — we solve it explicitly.
2. Take logs to free the exponent. $-(x^2+y^2)=\ln 0.5\Rightarrow x^2+y^2=-\ln 0.5=\ln 2$. Why the logarithm? The unknown is trapped in an exponent; $\ln$ is the tool that undoes $e^{(\cdot )}$, turning "$e$-to-the-something $=c$" into "something $=\ln c$". No other operation isolates the exponent.
3. Read off the radius. $x^2+y^2=\ln 2\approx 0.6931$ is a circle of radius $r=\sqrt{\ln 2}\approx 0.8326$ centered at the origin. Why this step? $x^2+y^2=r^2$ is the circle definition — so every level of a Gaussian bump is a circle, as the figure's red ring shows.
4. Degenerate top. The maximum of $f$ is at the origin where $f(0,0)=e^0=1$. The level $c=1$ solves $x^2+y^2=0$ — a single point, not a curve. Levels $c>1$ give the empty set (no contour drawn). Why this step? This is the limiting case: as $c\to$ the peak, the circle shrinks to a dot; beyond the peak nothing exists to slice.

Verify: $r=\sqrt{\ln 2}=\mathrm{0.83255...}$; check $e^{-r^2}=e^{-\ln 2}=1/2=0.5$ ✔. Max value $f(0,0)=1$; level $c=1$ is the single point $(0,0)$; $c>1$ empty ✔.

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Ex 7 — imshow orientation flip (cell C7)

Forecast: Guess whether row 0 sits at the top or bottom by default.

1. Row↔y rule. imshow maps Z[row, col] with row = y, col = x (parent trap #2). Value $30$ is at Z[1,0] → row $1$, column $0$. Why this step? We must translate array indices into screen position before we can say "top" or "bottom".
2. Default origin="upper". Row 0 is drawn at the top, rows increase downward. So row 1 (the $30/40$ row) is at the bottom, column 0 on the left → $30$ appears bottom-left. Why this step? Image files store row 0 at the top; that's the default and why images look "right" but math matrices look flipped.
3. origin="lower". Row 0 is drawn at the bottom, rows increase upward. Now row 1 is at the top, column 0 left → $30$ appears top-left. Why this step? origin="lower" makes increasing row = increasing $y$ (upward), which matches how we plot math where $y$ grows up.

Verify: With origin="upper", Z[1,0]=30 is bottom-left; flipping to origin="lower" moves it top-left — the two images are vertical mirror-flips of each other, exactly one row-reversal apart ✔.

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Ex 8 — imshow extent mapping with negatives (cell C8)

Forecast: Guess whether cell [0,0] lands exactly at the corner $(-3,-3)$ or slightly inside.

1. Cell width in real units. The x-range $[-3,3]$ spans $6$ units across $100$ columns, so each cell is $\Delta =6/100=0.06$ wide (same for y). Why this step? extent stretches the whole array to fill $[-3,3]$; each of 100 cells gets an equal slice.
2. Center of the first cell. Column 0 occupies real x from $-3$ to $-3+0.06=-2.94$; its center is $-3+0.06/2=-2.97$. With origin="lower", row 0 is at the bottom so its y-center is also $-2.97$. So Z[0,0] center $\to (-2.97,-2.97)$. Why this step? Pixel centers, not edges, are the sample locations; the edge is at $-3$ but the center sits half a cell in.
3. Center of the last cell. Column 99 spans x from $-3+99\cdot 0.06=2.94$ to $3.00$; center $=2.97$. Same for row 99 → Z[99,99] center $\to (2.97,2.97)$. Why this step? By symmetry the far corner mirrors the near corner; the grid is symmetric about the origin.

Verify: Cell width $0.06$; first center $-2.97$, last center $2.97$; the two centers are symmetric ($-2.97=-2.97$) and span $2.97-(-2.97)=5.94=99\cdot 0.06$ ✔. This uses the same even-spacing idea as NumPy meshgrid and broadcasting.

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Ex 9 — Real-world word problem (cell C9)

Forecast: Guess which of the six plots fits ordered time vs a pile of samples.

1. Temperature vs time → line. Time is an ordered axis and consecutive readings are connected as a trend, so ax.plot. Why this step? The parent decision table: "trend over an ordered axis → line". Connecting the dots is meaningful here (temperature between hours is continuous).
2. Reaction times → histogram. 500 independent samples where we care about the shape of the distribution, not order, so ax.hist with $B\approx \sqrt{500}\approx 22$ bins. Why this step? Table: "distribution/shape of one variable → histogram"; $\sqrt{n}$ is the parent's bin rule of thumb.
3. The mean. Mean $=\frac{528}{24}=22$ °C. A line plot shows the trend (when it peaked); a histogram would collapse the 24 ordered readings into value-buckets and lose the daily rhythm — so for time structure, line wins; for typical value, the mean/histogram wins. Why this step? Choosing the plot IS the question — each answers a different thing (WHEN vs HOW-OFTEN).

Verify: $\sqrt{500}=\mathrm{22.36...}$ → about 22 bins ✔. Mean $=528/24=22.0$ °C ✔ (units: degrees Celsius). Line for the 24-point ordered series, histogram for the 500-sample pile ✔.

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Ex 10 — Exam twist: meshgrid shape meets contour (cell C10)

Forecast: Guess whether X is $5\times 9$ or $9\times 5$ — the order trips almost everyone.

1. Grid shape. meshgrid(xs, ys) returns arrays of shape (len(ys), len(xs)) = (9, 5) — rows follow ys, columns follow xs (the parent's $X[i,j]=x_j$, $Y[i,j]=y_i$). So X.shape = Z.shape = (9, 5). Why this step? Contour needs Z shaped (len(y), len(x)); getting it backwards raises a shape error. This is the parent mistake "contour takes my 1D x and y" made concrete.
2. Number of contour lines. levels=[1, 4, 9] lists 3 explicit levels, so up to 3 contour curves are drawn (one per level that actually intersects the surface). Why this step? Passing a list (not an int) means "draw exactly these levels", unlike levels=8 which means "pick 8 nice levels automatically".
3. A point on level $c=4$. Solve $x^2+y=4$. Pick a grid $x$ value: $x=0\Rightarrow y=4$; the point $(0,4)$ is on the surface's grid (since $y=4$ is one of ys) and satisfies $Z=0+4=4$. Another: $x=2\Rightarrow y=0$, and $(2,0)$ gives $Z=4+0=4$. Why this step? The level set is literally the solution set of $f(x,y)=c$; we exhibit members of it to prove the line exists.

Verify: meshgrid(xs, ys) → shape (9,5) ✔ (9 rows, 5 cols). levels=[1,4,9] → 3 lines ✔. On $c=4$: $Z(0,4)=0^2+4=4$ ✔ and $Z(2,0)=2^2+0=4$ ✔.

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Recall Rapid self-test

Bin width for range $[0,10]$, $B=5$? ::: $h=2$. Does a value exactly on a bin's left edge go left or right? ::: Right (bins are $[\cdot ,\cdot )$ half-open). Density heights sum-of-areas must equal? ::: $1$ (since $\sum c_j=n$). Radius of $e^{-(x^2+y^2)}=0.5$ contour? ::: $\sqrt{\ln 2}\approx 0.83$. meshgrid(xs, ys) gives X of what shape? ::: (len(ys), len(xs)). origin="lower" vs "upper" — which matches math matrices? ::: "lower" (row 0 at bottom, $y$ up). Bar at category $k$, width $w$ — left edge? ::: $k-w/2$.

See also: Matplotlib Figure vs Axes object model · Colormaps and perceptual uniformity (viridis) · Image arrays and pixel coordinates · 3D surface plots (plot_surface) · Seaborn — statistical plotting wrappers.