d3.tsvParseRows(string[, row]) Equivalent to dsvFormat("\t").parseRows.

d3.tsvParse(string[, row]) Equivalent to dsvFormat("\t").parse.

d3.tsvFormatRows(rows) Equivalent to dsvFormat("\t").formatRows.

d3.tsvFormat(rows[, columns]) Equivalent to dsvFormat("\t").format.

d3.tsv(url[[, row], callback]) Creates a request for the TSV file at the specified url with the default mime type text/tab-separated-values. An optional row conversion function may be specified to map and filter row objects to a more-specific representation; see dsv.parse for details. For example: function row(d) { return { year: new Date(+d.Year, 0, 1), // convert "Year" column to Date make: d.Make, model: d.Model, length: +d.Length // convert "Length" column to number };

d3.treemapSquarify(node, x0, y0, x1, y1) Implements the squarified treemap algorithm by Bruls et al., which seeks to produce rectangles of a given aspect ratio.

d3.treemapSliceDice(node, x0, y0, x1, y1) If the specified node has odd depth, delegates to treemapSlice; otherwise delegates to treemapDice.

d3.treemapSlice(node, x0, y0, x1, y1) Divides the rectangular area specified by x0, y0, x1, y1 vertically according the value of each of the specified node’s children. The children are positioned in order, starting with the top edge (y0) of the given rectangle. If the sum of the children’s values is less than the specified node’s value (i.e., if the specified node has a non-zero internal value), the remaining empty space will be positioned on the bottom edge (y1) of the given rectangle.

d3.treemapResquarify(node, x0, y0, x1, y1) Like d3.treemapSquarify, except preserves the topology (node adjacencies) of the previous layout computed by d3.treemapResquarify, if there is one and it used the same target aspect ratio. This tiling method is good for animating changes to treemaps because it only changes node sizes and not their relative positions, thus avoiding distracting shuffling and occlusion. The downside of a stable update, however, is a suboptimal layout for subsequent upda

d3.treemapDice(node, x0, y0, x1, y1) Divides the rectangular area specified by x0, y0, x1, y1 horizontally according the value of each of the specified node’s children. The children are positioned in order, starting with the left edge (x0) of the given rectangle. If the sum of the children’s values is less than the specified node’s value (i.e., if the specified node has a non-zero internal value), the remaining empty space will be positioned on the right edge (x1) of the given rectangle.