25:["$","title",null,{"children":"Benford Distribution | Maya Graham | Observable"}] 49:T8e2, ### Explanation of Newcomb-Benford Law Newcomb-Benford's (NB) Law states the first digit of numbers in real-life data sets is governed by ${tex`p_i = \log_{10}\left (1 + \frac{1}{i}\right)`} where ${tex`p_i`} is the probability of the digit ${tex`i \in [1...9]`}. The following table shows the expected probability for each digit. | ${tex`i`} | ${tex`p_i`} | |---|----------| | 1 | 0.30103 | | 2 | 0.17609 | | 3 | 0.12494 | | 4 | 0.09691 | | 5 | 0.079181 | | 6 | 0.066947 | | 7 | 0.057992 | | 8 | 0.051153 | | 9 | 0.045757 | This "Law" in data occurs frequently enough that it has been used to test for accounting fraud, government corruption, and the validity of scientific results, such as COVID-19 data. Not all data have this distribution, for NB's law to hold data should: - **have sufficient span** either a complete or several orders of magnitude in the data's span. - **have large enough sample size** random chance could make a few first digits non-NB. If the data set is small enough (say 25 or less), this could have a significant impact on statistical tests for NB distributions. - **be right skewed** often this is associated with some combinatory or self-organized aspect of the data. Right skew generally increases the span, returning us to the first expectation. - **not be human assigned** serial numbers or telephone numbers would not work. Likewise, numbers that have been manipulated by humans to serve some purpose will not work. Identifying such numbers is the point of this exercise. - **not be bound** the data should not have a hard maximum or minimum imposed by the physical setting. For example, daily temperatures are not Benford distributed. One what to think of data having these characteristics is that the data have some underlying, exponential process operating generate it. This would mean that random samples from the data would stretch across orders of magnitude, but be the samples would be evenly distributed across several orders of magnitude. As such, we see why NB's Law emerges from data sets having the above characteristics, by considering logarithmic axes drawn on top of linear axes. Can you see the red lines on the log scale are at exactly the points give by ${tex`p_i = \log_{10}\left (1 + \frac{1}{i}\right)`}?4a:T811,Plot.plot({ height: 500, width: 700, marginTop: 60, marginBottom: 50, style: {fontSize: "15px"}, x: { label: "Leading Digit", type: "band" }, y: { percent: true, label: "Percent of Total Occurrences" }, color: { legend: true, domain: ["Actual Data", "Benford Expected"], range: ["mediumaquamarine", "black"] }, title: "Leading Digit Distribution vs Expected Benford Distribution", caption: "Bars show the frequency of digits found in the data set and the labled line shows the percent expected under a Benford distribution", marks: [ Plot.barY( purchasesDigit1.filter(d => d.digit_1 >= 1 && d.digit_1 <= 9), Plot.groupX( { y: "proportion", title: (digit) => { const total = purchasesDigit1.filter(d => d.digit_1 >= 1 && d.digit_1 <= 9).length; return `${((digit.length / total) * 100).toFixed(1)}%`; } }, { x: "digit_1", tip: true, fill: "mediumaquamarine" } ) ), Plot.lineY([ { d: 1, p: 0.301 }, { d: 2, p: 0.176 }, { d: 3, p: 0.125 }, { d: 4, p: 0.097 }, { d: 5, p: 0.079 }, { d: 6, p: 0.067 }, { d: 7, p: 0.058 }, { d: 8, p: 0.051 }, { d: 9, p: 0.046 } ], { x: "d", y: "p", stroke: "cadetblue", strokeWidth: 2 }), Plot.dot([ { d: 1, p: 0.301 }, { d: 2, p: 0.176 }, { d: 3, p: 0.125 }, { d: 4, p: 0.097 }, { d: 5, p: 0.079 }, { d: 6, p: 0.067 }, { d: 7, p: 0.058 }, { d: 8, p: 0.051 }, { d: 9, p: 0.046 } ], { x: "d", y: "p", fill: "black", r: 4 }), Plot.text([ { d: 1, p: 0.301 }, { d: 2, p: 0.176 }, { d: 3, p: 0.125 }, { d: 4, p: 0.097 }, { d: 5, p: 0.079 }, { d: 6, p: 0.067 }, { d: 7, p: 0.058 }, { d: 8, p: 0.051 }, { d: 9, p: 0.046 } ], { x: "d", y: "p", text: (item) => `${(item.p * 100).toFixed(1)}%`, dy: -20, fontSize: 13, fontWeight: "bold", fill: "black" }) ] })4b:T1c5c,// Place code to create your plot here! { // ============================================================ // CONSTANTS // ============================================================ const RANDOM_LABEL = "Random"; const SMALL_SAMPLE_THRESHOLD = 50; // Non-uniform slider steps: 1s, then 10s, 100s, 1000s, 10000s const SLIDER_STEPS = buildSteps([ [1, 10, 1], [10, 100, 10], [100, 1000, 100], [1000, 10000, 1000], [10000, 100000, 10000] ]); function buildSteps(ranges) { const values = new Set(); for (const [start, end, step] of ranges) { for (let v = start; v <= end; v += step) values.add(v); } return [...values].sort((a, b) => a - b); } // ============================================================ // BENFORD MATH // ============================================================ // Extracts the leading digit from a value. Handles plain numbers, // numeric strings, and objects like {digit_1: 4}. function getFirstDigit(x) { if (x !== null && typeof x === "object") x = x.digit_1; return parseInt(String(x)[0]); } // Converts a raw column (which may contain blanks/NaN/strings) into // a clean array of finite numbers. function cleanNumbers(values) { return values .map(v => (typeof v === "string" ? +v : v)) .filter(v => typeof v === "number" && Number.isFinite(v)); } // Turns an array of values into first-digit probabilities: [{digit, p_i}, ...] function digitFrequency(values) { const digits = values.map(getFirstDigit).filter(d => d >= 1 && d <= 9); const counts = d3.rollup(digits, v => v.length, d => d); return d3.range(1, 10).map(digit => ({ digit, p_i: (counts.get(digit) ?? 0) / digits.length })); } // The d* goodness-of-fit statistic: 0 = perfectly Benford, 1 = maximally not. function dStar(observedFreq) { const sumSquares = d3.sum( Benford_distribution.map((b, i) => (b.p_i - observedFreq[i].p_i) ** 2) ); return Math.sqrt(sumSquares) / 1.036061; } function verdictFor(d) { if (d < 0.19) return "consistent with Benford's Law."; if (d < 0.25) return "some deviation from Benford — worth a second look."; return "substantial deviation — potentially non-Benford or manipulated data."; } // Draws n random values spread across `span` orders of magnitude, // then takes each one's leading digit. function randomSample(n) { return d3.range(n).map( () => parseInt(Math.pow(10, d3.randomUniform()() * span).toString()[0]) ); } // ============================================================ // DATASETS // ============================================================ const columns = { "Mystery A": cleanNumbers(mystery.map(d => d.A)), "Mystery B": cleanNumbers(mystery.map(d => d.B)), "Mystery C": cleanNumbers(mystery.map(d => d.C)), "Mystery D": cleanNumbers(mystery.map(d => d.D)), "Mystery E": cleanNumbers(mystery.map(d => d.E)), "Mystery F": cleanNumbers(mystery.map(d => d.F)), "Actual Purchases": purchasesDigit1 }; // Label shown on each radio button, e.g. "Mystery A (n=114)" const labelWithCount = name => `${name} (n=${columns[name].length})`; // ============================================================ // UI: dataset radio buttons, grouped into three rows // ============================================================ let selectedDataset = RANDOM_LABEL; function makeRadioRow(names) { const row = html`

`; for (const name of names) { const label = name === RANDOM_LABEL ? RANDOM_LABEL : labelWithCount(name); const id = `dataset-${name.replace(/\W+/g, "-")}`; const option = html` ${label} `; option.querySelector("input").checked = name === selectedDataset; option.querySelector("input").addEventListener("change", () => { selectedDataset = name; render(); }); row.append(option); } return row; } const row1 = makeRadioRow(["Mystery A", "Mystery B", "Mystery C"]); const row2 = makeRadioRow(["Mystery D", "Mystery E", "Mystery F"]); const row3 = makeRadioRow(["Actual Purchases", RANDOM_LABEL]); // ============================================================ // UI: sample-size slider (only shown when "Random" is selected) // ============================================================ // Slider position is an INDEX into SLIDER_STEPS, not the n value itself. // This keeps arrow-key presses moving one step at a time even though // the underlying n values jump unevenly (1, 2, ...9, 10, 20, ...). const slider = html``; const bubble = html`

`; const sliderRow = html`

${bubble}${slider}

`; const currentN = () => SLIDER_STEPS[+slider.value]; function moveBubbleToThumb() { const pct = +slider.value / (SLIDER_STEPS.length - 1); const trackWidth = slider.offsetWidth || 250; const thumbRadius = 8; bubble.style.left = `${thumbRadius + pct * (trackWidth - thumbRadius * 2)}px`; bubble.textContent = `n = ${currentN()}`; } slider.addEventListener("input", () => { moveBubbleToThumb(); render(); }); // ============================================================ // CHART // ============================================================ const chartArea = html`

`; function render() { const isRandom = selectedDataset === RANDOM_LABEL; sliderRow.style.display = isRandom ? "" : "none"; const values = isRandom ? randomSample(currentN()) : columns[selectedDataset]; const freq = digitFrequency(values); const d = dStar(freq); const title = isRandom ? `${RANDOM_LABEL} (n=${values.length})` : selectedDataset; const plot = Plot.plot({ title: `${title}, d* = ${d.toFixed(3)}`, marginLeft: 80, marginBottom: 60, marginTop: 40, x: {label: "First digit", domain: d3.range(1, 10)}, y: {label: "Probability", grid: true, domain: [0, 0.5], tickSize: 6}, marks: [ Plot.barY(freq, {x: "digit", y: "p_i", fill: "mediumaquamarine"}), Plot.lineY(Benford_distribution, {x: "digit", y: "p_i", stroke: "cadetblue", strokeWidth: 2}), Plot.dot(Benford_distribution, {x: "digit", y: "p_i", fill: "black", r: 4}) ] }); chartArea.innerHTML = ""; chartArea.append(plot); } // ============================================================ // INITIAL RENDER // ============================================================ moveBubbleToThumb(); render(); return html`

${row1}${row2}${row3}${sliderRow}${chartArea}

`; }4c:T92b,Plot.plot({ height: 400, width: 700, marginTop: 40, marginBottom: 50, style: {fontSize: "15px"}, x: { label: "Leading Digit", type: "band" }, y: { percent: true, label: "Percent of Total Occurrences" }, // 1. Define your color scale and enable the legend color: { domain: ["Actual Data", "Expected (Benford)"], range: ["steelblue", "crimson"], legend: true, }, title: "Distribution of Leading Digits Compared to Expected Benford Distribution", marks: [ // Your existing bar mark - updated fill to use the color domain key Plot.barY( purchasesDigit1.filter(d => d.digit_1 >= 1 && d.digit_1 <= 9), Plot.groupX( { y: "proportion", title: (digit) => { const total = purchasesDigit1.filter(d => d.digit_1 >= 1 && d.digit_1 <= 9).length; return `${((digit.length / total) * 100).toFixed(1)}%`; } }, { x: "digit_1", tip: true, fill: () => "Actual Data" // Maps directly to steelblue via the color domain } ) ), // Line mark - updated stroke to use the color domain key Plot.lineY([ { d: 1, p: 0.301 }, { d: 2, p: 0.176 }, { d: 3, p: 0.125 }, { d: 4, p: 0.097 }, { d: 5, p: 0.079 }, { d: 6, p: 0.067 }, { d: 7, p: 0.058 }, { d: 8, p: 0.051 }, { d: 9, p: 0.046 } ], { x: "d", y: "p", stroke: () => "Expected (Benford)", // Maps directly to crimson via the color domain strokeWidth: 2 }), // Dot mark - updated fill to use the color domain key Plot.dot([ { d: 1, p: 0.301 }, { d: 2, p: 0.176 }, { d: 3, p: 0.125 }, { d: 4, p: 0.097 }, { d: 5, p: 0.079 }, { d: 6, p: 0.067 }, { d: 7, p: 0.058 }, { d: 8, p: 0.051 }, { d: 9, p: 0.046 } ], { x: "d", y: "p", fill: () => "Expected (Benford)", r: 4 }), // Text labels remain unchanged Plot.text([ { d: 1, p: 0.301 }, { d: 2, p: 0.176 }, { d: 3, p: 0.125 }, { d: 4, p: 0.097 }, { d: 5, p: 0.079 }, { d: 6, p: 0.067 }, { d: 7, p: 0.058 }, { d: 8, p: 0.051 }, { d: 9, p: 0.046 } ], { x: "d", y: "p", text: (item) => `${(item.p * 100).toFixed(1)}%`, dy: -10, fontSize: 11, fontWeight: "bold", fill: "crimson" }) ] }) 4d:T795,Plot.plot({ height: 500, width: 700, marginTop: 60, marginBottom: 50, style: {fontSize: "15px"}, x: { label: "Leading Digit", type: "band" }, y: { percent: true, label: "Percent of Total Occurrences" }, title: "Leading Digit Distribution vs Expected Benford Distribution", caption: "Bars show the frequency of digits found in the data set and the labled line shows the percent expected under a Benford distribution", marks: [ Plot.barY( purchasesDigit1.filter(d => d.digit_1 >= 1 && d.digit_1 <= 9), Plot.groupX( { y: "proportion", title: (digit) => { const total = purchasesDigit1.filter(d => d.digit_1 >= 1 && d.digit_1 <= 9).length; return `${((digit.length / total) * 100).toFixed(1)}%`; } }, { x: "digit_1", tip: true, fill: "mediumaquamarine" } ) ), Plot.lineY([ { d: 1, p: 0.301 }, { d: 2, p: 0.176 }, { d: 3, p: 0.125 }, { d: 4, p: 0.097 }, { d: 5, p: 0.079 }, { d: 6, p: 0.067 }, { d: 7, p: 0.058 }, { d: 8, p: 0.051 }, { d: 9, p: 0.046 } ], { x: "d", y: "p", stroke: "cadetblue", strokeWidth: 2 }), Plot.dot([ { d: 1, p: 0.301 }, { d: 2, p: 0.176 }, { d: 3, p: 0.125 }, { d: 4, p: 0.097 }, { d: 5, p: 0.079 }, { d: 6, p: 0.067 }, { d: 7, p: 0.058 }, { d: 8, p: 0.051 }, { d: 9, p: 0.046 } ], { x: "d", y: "p", fill: "black", r: 4 }), Plot.text([ { d: 1, p: 0.301 }, { d: 2, p: 0.176 }, { d: 3, p: 0.125 }, { d: 4, p: 0.097 }, { d: 5, p: 0.079 }, { d: 6, p: 0.067 }, { d: 7, p: 0.058 }, { d: 8, p: 0.051 }, { d: 9, p: 0.046 } ], { x: "d", y: "p", text: (item) => `${(item.p * 100).toFixed(1)}%`, dy: -20, fontSize: 13, fontWeight: "bold", fill: "black" }) ] }) 4e:T603,Plot.plot({ height: 400, width: 700, marginTop: 40, marginBottom: 50, style: {fontSize: "15px"}, x: { label: "Leading Digit", type: "band" }, y: { percent: true, label: "Percent of Total Occurrences" }, title: "Leading Digit Distribution vs Expected Benford Distribution", marks: [ // Your existing bar mark Plot.barY( purchasesDigit1.filter(d => d.digit_1 >= 1 && d.digit_1 <= 9), Plot.groupX( { y: "proportion", title: (digit) => { const total = purchasesDigit1.filter(d => d.digit_1 >= 1 && d.digit_1 <= 9).length; return `${((digit.length / total) * 100).toFixed(1)}%`; } }, { x: "digit_1", tip: true, fill: "mediumaquamarine" } ) ), // Line mark matching numerical values Plot.lineY([ { d: 1, p: 0.301 }, { d: 2, p: 0.176 }, { d: 3, p: 0.125 }, { d: 4, p: 0.097 }, { d: 5, p: 0.079 }, { d: 6, p: 0.067 }, { d: 7, p: 0.058 }, { d: 8, p: 0.051 }, { d: 9, p: 0.046 } ], { x: "d", y: "p", stroke: "darkslategrey", strokeWidth: 2 }), // Dot mark matching numerical values Plot.dot([ { d: 1, p: 0.301 }, { d: 2, p: 0.176 }, { d: 3, p: 0.125 }, { d: 4, p: 0.097 }, { d: 5, p: 0.079 }, { d: 6, p: 0.067 }, { d: 7, p: 0.058 }, { d: 8, p: 0.051 }, { d: 9, p: 0.046 } ], { x: "d", y: "p", fill: "darkslategrey", r: 4, tip: true }) ] })4f:T4c8,### Testing for Benford Distributions Give a data set that may be Bedford, the first digits can be evaluated to estimate the probability of each digit's value. Call that data set ${tex`p^\dagger`}. In addition to visual inspection, the ${tex`d^*`} statistic is frequently used ${tex`d^* = \frac{1}{ 1.03606 }\sqrt{ \sum_{i=1}^9 (p_i-p^\dagger_i)^2} `}. Where the denominator is just the maximum difference between Benford distribution of first digits and some data set. This makes ${tex`d^* \in`} [0,1] where ${tex`d^* =`} 0 is maximally Benford and ${tex`d^* =`} 1 is maximally non-Benford. Provided there are sufficient samples, values of ${tex`d^* > `} 0.25 might be considered fraudulent, and values over 0.19 should attract additional scrutiny. However, questions about the span of the data and the number of samples make firm guidelines difficult to state. ### Further reading [An introduction](https://rss.onlinelibrary.wiley.com/doi/pdf/10.1111/j.1740-9713.2016.00919.x) ## Something to try Below I've uploaded a CSV file. It consists of 6 columns of different length. They are all drawn from data you may find interesting. In your groups, develop a method of determining if each column is Bedford distributed. 50:[{"id":"0786be72140ead175fb2d952e668c209b605c71350d9dd66b9f0af6ca796b6f7ac0bfde12394fa9e3d1f576ccd606e1772e07e001d2ab1878f452522a16e9784","href":"https://static.observableusercontent.com/files/0786be72140ead175fb2d952e668c209b605c71350d9dd66b9f0af6ca796b6f7ac0bfde12394fa9e3d1f576ccd606e1772e07e001d2ab1878f452522a16e9784","name":"2023 Purchases - digit1.csv","type":"text/csv","lastModified":1729629855746.169,"size":4230},{"id":"82bcda9673e15d725fa2d123ee12ae14f6d4255ce997e832654edcc7f2314c44715b2da3c80115c4158f6eec3f5c9b6321bc23a552a06c5d0fc5f9c39fa2b6e6","href":"https://static.observableusercontent.com/files/82bcda9673e15d725fa2d123ee12ae14f6d4255ce997e832654edcc7f2314c44715b2da3c80115c4158f6eec3f5c9b6321bc23a552a06c5d0fc5f9c39fa2b6e6","name":"Mystery@1.csv","type":"text/csv","lastModified":1729697438749.061,"size":5297}] 29:{"id":"7f7ab4be36f966fe","slug":null,"title":"Benford Distribution","thumbnail":"812248492f0139b1d0dda6f00534bb5954673f2377eebc6fd8d11471e344fef2","access_level":"public","create_time":"$D2026-09-14T17:14:14.601Z","update_time":"$D2026-09-16T17:32:33.565Z","view_time":null,"first_public_time":"$D2026-09-16T17:16:49.776Z","has_unpublished_edits":null,"fork":{"id":"dbd2bc5cf33a1bd1","slug":null,"title":"Benford Distribution","owner":{"id":"518a19f15bc65c30","login":"jesse-johnson-ws","name":"Jesse Johnson's Workspace"}},"stars":0,"starred":false,"owner":{"type":"team","id":"81ae52680c7448b3","login":"maya-graham","name":"Maya Graham","avatar_url":"https://avatars-next.observableusercontent.com/2b31b2d3687ff71f3b77c32bbce4334195892043b8fba121bec8886284dbe305","tier":"starter_2024","bio":"","following":false},"version":477,"publish_version":null,"role":null,"authored":false,"authors":[{"type":"individual","id":"f2d6a0cdec3e6ab6","login":"maya-graham","name":"Maya Graham","avatar_url":"https://avatars-next.observableusercontent.com/2b31b2d3687ff71f3b77c32bbce4334195892043b8fba121bec8886284dbe305","tier":null,"bio":"","following":false}],"members":[{"type":"team","id":"81ae52680c7448b3","login":"maya-graham","name":"Maya Graham","avatar_url":"https://avatars-next.observableusercontent.com/2b31b2d3687ff71f3b77c32bbce4334195892043b8fba121bec8886284dbe305","tier":"starter_2024","role":"editor"}],"comments":[],"custom_thumbnail":false,"trashed":false,"lapsed":null,"subscribed":null,"forks":0,"collection":null,"type":"notebook","body":{"titling":"auto","title":"Benford Distribution","cells":[{"id":0,"value":"# Benford Distribution\nThe objective of this in-class exercise is to visually compare the first digit frequencies of a set of randomly generated data to the Benford distribution. In groups, you will work to achieve the most informative and easily understood visualization possible.","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"","database":"$undefined"},{"id":272,"value":"$49","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"","database":"$undefined"},{"id":274,"value":"Plot.plot({\n x: { type: \"log\", domain: [1, 10], grid: true }\n})","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":278,"value":"Plot.plot({\n x: {\n type: \"linear\",\n domain: [0, 1],\n grid: true\n },\n marks: [\n Plot.ruleX(\n d3\n .range(1, 10)\n .map((x) => Math.log10(1 + 1 / x))\n .map(\n (\n (sum) => (value) =>\n (sum += value)\n )(0)\n ),\n { stroke: \"red\" }\n )\n ]\n})","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":301,"value":"## Assignment\nBased on your understanding of the Newcomb-Benford law develop quantitative and graphical methods for testing a set of numbers for NB. Your graphs should do two things: 1) show the distribution of first digits in relation to the values anticipated by NB. 2) provide a single, scalar valued statistical summary of the differences.\n\nSee below for a formula for the statistical summary.\n\nI've provided some conveniences variables below, as javascript.\n* `Benford_distribution` provides the first digit probabilities for the distribution.\n* `Random_distribution` provides a random distribution of first digits.\n* `bins` counts how many of those randomly sampled first digits are each of the nine digits.\n\nThe visual your group creates should show the Benford predictions and the results of the random sampled numbers. Fork this page and get going!\n\n*** You are on your honor to not look at the other forks! I want to see original ideas for how to do this. ***","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":464,"value":"$4a","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":473,"value":"$4b","pinned":true,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":460,"value":"$4c","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":437,"value":"$4d","pinned":true,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":424,"value":"$4e","pinned":true,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":320,"value":"Plot.plot({\n height: 400,\n width: 700,\n marginTop: 40,\n marginBottom: 50,\n style:{fontSize: \"15px\"},\n x: {\n label: \"Leading Digit\",\n },\n y: {\n percent: true,\n label: \"Percent of Total Occurrences\"\n },\n title: \"Distribution of Leading Digits Compared to Expected Benford Distribution\",\n //caption:\n \n \n \n marks: [\n Plot.barY(\n purchasesDigit1.filter(d => d.digit_1 >= 1 && d.digit_1 <= 9), \n Plot.groupX(\n { \n y: \"proportion\",\n //title: (digit) => `${((digit.length/length) * 100).toFixed(1)}%`,\n title: (digit) => {\n const total = purchasesDigit1.filter(d => d.digit_1 >= 1 && d.digit_1 <= 9).length;\n return `${((digit.length / total) * 100).toFixed(1)}%`;\n }\n }, \n {\n x: \"digit_1\",\n tip: true,\n fill: \"steelblue\"\n }\n )\n )\n ]\n})","pinned":true,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":318,"value":"Plot.plot({\n marks: [\n Plot.rectY(olympians, Plot.binX({y: \"count\"}, {x: \"weight\"})),\n Plot.ruleY([0])\n ]\n})","pinned":true,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":194,"value":"// Place code to create your plot here!","pinned":true,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":5,"value":"Benford_distribution = d3\n .range(1, 10)\n .map((x) => ({ digit: x, p_i: Math.log10(1 + 1 / x) }))","pinned":true,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":290,"value":"// This is the number of samples that will be used to compute randomly generated Benford distributions see how many samples you need to get really good vs somewhat noisy distributions.\nnumber_of_samples = 10000","pinned":true,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":292,"value":"// This is the number of orders of magnitude the random_distribution is over. See how 'Benford' this is for shorter spans vs longer ones\nspan = 3","pinned":true,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":47,"value":"random_distribution = d3.range(number_of_samples).map((x) => ({\n sample: x,\n value: parseInt(Math.pow(10, d3.randomUniform().call() * span).toString()[0])\n}))","pinned":true,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":51,"value":"function getFirstDigit(x) {\n // Note this wont work if x < 1\n return parseInt(x.toString[0]);\n}","pinned":true,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":137,"value":"bins = d3\n .bin()\n .thresholds(9)\n .value((d) => d.value)(random_distribution)","pinned":true,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":264,"value":"## Test Data\nBelow I've uploaded the first digit of all of my 2023 credit card purchases. What do you think? Benford? Try it out! Use the plot technique you developed above.","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"","database":"$undefined"},{"id":261,"value":"purchasesDigit1 = await FileAttachment(\"2023 Purchases - digit1.csv\").csv({typed: \"auto\"})","pinned":true,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"purchasesDigit1","database":"$undefined"},{"id":298,"value":"purchasesDigit1[0].digit_1","pinned":true,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":287,"value":"$4f","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"","database":"$undefined"},{"id":296,"value":"mystery = await FileAttachment(\"Mystery@1.csv\").csv({typed: \"auto\"})","pinned":true,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"mystery","database":"$undefined"}],"files":"$W50","stdlib":"1"}} 24:null