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General Update Pattern, I
license: gpl-3.0

Note: d3-selection 1.4 introduced selection.join which greatly simplifies the general update pattern! This series of examples is useful for understanding how joins work in D3, but I recommend you use selection.join in practice.

This example demonstrates D3’s general update pattern, where a data-join is followed by operations on the enter, update and exit selections. Entering elements are shown in green, while updating elements are shown in black. Exiting elements are removed immediately, so they're invisible.

This example does not use a key function for the data-join, so elements may change their associated letter. Entering elements are always added to the end: when the new data has more letters than the old data, new elements are entered to display the new letters. Likewise, exiting letters are always removed from the end when the new data has fewer letters than the old data.

Next: Key Functions

<!DOCTYPE html>
<meta charset="utf-8">
text {
font: bold 48px monospace;
.enter {
fill: green;
.update {
fill: #333;
<svg width="960" height="500"></svg>
<script src=""></script>
var alphabet = "abcdefghijklmnopqrstuvwxyz".split("");
var svg ="svg"),
width = +svg.attr("width"),
height = +svg.attr("height"),
g = svg.append("g").attr("transform", "translate(32," + (height / 2) + ")");
function update(data) {
// Join new data with old elements, if any.
var text = g.selectAll("text")
// Update old elements as needed.
text.attr("class", "update");
// Create new elements as needed.
// After merging the entered elements with the update selection,
// apply operations to both.
.attr("class", "enter")
.attr("x", function(d, i) { return i * 32; })
.attr("dy", ".35em")
.text(function(d) { return d; });
// Remove old elements as needed.
// The initial display.
// Grab a random sample of letters from the alphabet, in alphabetical order.
d3.interval(function() {
.slice(0, Math.floor(Math.random() * 26))
}, 1500);
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fangmdu commented Dec 14, 2012

//code line 68
.slice(0,Math.floor(Math.random()* 26)+1)

random from 1- 26.

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Jonahss commented Apr 26, 2013

I was reading this example which you linked to today in your post on Selections.
The way I understand it, it doesn't work as specified. The black updated letters in each iteration are not letter which were present in the previous step.

I think it's because the array is shuffled, but data is connected to nodes based on the index within the alphabet array?

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