假设我想要一个字符串,比如“123”来填充给定的矩形,如下所示:
Show[Plot[x, {x, 0, 1}],
Graphics[{EdgeForm[Thick], Yellow, Rectangle[{.1, .5}, {.4, .9}]}],
Graphics[Text[Style["123", Red, Bold, 67], {.1, .5}, {-1, -1}]]]
但我手动调整那里的字体大小(67),以便填满矩形。 如何使任意字符串填充任意矩形?
答案 0 :(得分:8)
我认为这是一个众所周知的难题。我能找到的最佳答案is from John Fultz.
TextRect[text_, {{left_, bottom_}, {right_, top_}}] :=
Inset[
Pane[text, {Scaled[1], Scaled[1]},
ImageSizeAction -> "ResizeToFit", Alignment -> Center],
{left, bottom}, {Left, Bottom}, {right - left, top - bottom}]
Show[
Plot[x, {x, 0, 1}],
Graphics[{
{EdgeForm[Thick], Yellow, Rectangle[{.1, .5}, {.4, .9}]},
TextRect[Style["123", Red, Bold], {{.1, .5}, {.4, .9}}]
}]
]
答案 1 :(得分:2)
这是另一种将文本转换为映射到多边形的纹理的方法。这具有拉伸文本以适应该区域的特征(因为它不再是文本了。)
Show[Plot[x, {x, 0, 1}],
Graphics[{EdgeForm[Thick], Yellow, Rectangle[{.1, .5}, {.4, .9}]}],
Graphics[{Texture[ImageData[
Rasterize[Style["123", Red, Bold], "Image", RasterSize -> 300,
Background -> None]]],
Polygon[{{0.1, 0.5}, {0.4, 0.5}, {0.4, 0.9}, {0.1, 0.9}},
VertexTextureCoordinates -> {{0, 0}, {1, 0}, {1, 1}, {0, 1}}]}]]
作为便于比较的功能:
(* Render string/style s to fill a rectangle with left/bottom corner {l,b} and
right/top corner {r,t}. *)
textrect[s_, {{l_,b_},{r_,t_}}] := Graphics[{
Texture[ImageData[Rasterize[s, "Image", RasterSize->300, Background->None]]],
Polygon[{{l,b}, {r,b}, {r,t}, {l,t}},
VertexTextureCoordinates->{{0,0},{1,0},{1,1},{0,1}}]}]
答案 2 :(得分:1)
当Plot不存在时,建议的解决方案不起作用,我使用PlotRange选项来解决它。我把它包裹在一个函数中;不透明度,文字颜色等;应该做成选择;
textBox[text_, color_, position_: {0, 0}, width_: 2, height_: 1] :=
Graphics[{
{
color, Opacity[.1],
Rectangle[position, position + {width, height},
RoundingRadius -> 0.1]
}
,
Inset[
Pane[text, {Scaled[1], Scaled[1]},
ImageSizeAction -> "ResizeToFit", Alignment -> Center],
position, {Left, Bottom}, {width, height}]
}, PlotRange ->
Transpose[{position, position + {width, height}}]];