Skip to main content

bugs - How to improve the recognition quality when TextRecognize work on single character


Bug introduced in 11.0 and persisting through 11.3




From this answer, I doubt the capability to work on single character. So I give some test to verify this possibility. You can get my test imgs by this code


imgs = Binarize[
Import[#]] & /@ {"https://i.stack.imgur.com/PvuFe.png",
"https://i.stack.imgur.com/bXHyv.png",
"https://i.stack.imgur.com/6Uxpo.png"};


Note the TextRecognize[#, "Character"] & /@ imgs will get nothing. We can get a example from the documentation in Examples/Applications, that indicate the appropriate mask maybe can improve the performance to get a character, but I don't very like this method. Because it is hard to build a mask for characters "i","j" like


TextRecognize[#, 
Masking -> MorphologicalTransform[#, "BoundingBoxes", Infinity],
RecognitionPrior -> "Character"] & /@ imgs


{{H,1,O},{m,Y},{d,d}}






  • Any workaround that can improve the recognition quality when TextRecognize work on single character


Or



  • If we want to improve the recognition quality by mask, how to build correct mask?


I desire to make my this answer better by TextRecognize.



Answer



I felt that I miss some simple way to unite closely located components and finally I found it: ImageForestingComponents (thanks to this answer)!




  • It is unfortunate that a link to this function isn't included in the "See Also" drop-down list neither on the Docs page for ComponentMeasurements, nor MorphologicalComponents, nor MorphologicalTransform. That's why I wasn't able to find it quickly...


I'll show how it can be used on the most problematic case with letter "i" which is formed by two not connected clusters of points:


i = Import["https://i.stack.imgur.com/PvuFe.png"]


image



With horizontal radius 1 and vertical radius 6 we get a segmentation where our letter "i" is counted as a single component:



ImageForestingComponents[i, Automatic, {1, 6}] // Colorize


image



Using ComponentMeasurements we can get the bounding boxes of our characters dropping the background:


c = ComponentMeasurements[ImageForestingComponents[i, Automatic, {1, 6}], 
"BoundingBox", #"ConvexCoverage" < .9 &]



{2 -> {{66., 125.}, {79., 161.}}, 3 -> {{46., 61.}, {84., 98.}}}

HighlightImage[i, {Yellow, Rectangle @@@ c[[All, 2]]}]


image



TextRecognize accepts a set of Rectangle primitives as a Mask (it is documented under the Examples ► Options ► Masking sub-subsection):


TextRecognize[i, Masking -> Rectangle @@@ c[[All, 2]], RecognitionPrior -> "Character"]



{"i", "O"}

That's all. :^)


Comments

Popular posts from this blog

mathematical optimization - Minimizing using indices, error: Part::pkspec1: The expression cannot be used as a part specification

I want to use Minimize where the variables to minimize are indices pointing into an array. Here a MWE that hopefully shows what my problem is. vars = u@# & /@ Range[3]; cons = Flatten@ { Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; Minimize[{Total@((vec1[[#]] - vec2[[u[#]]])^2 & /@ Range[1, 3]), cons}, vars, Integers] The error I get: Part::pkspec1: The expression u[1] cannot be used as a part specification. >> Answer Ok, it seems that one can get around Mathematica trying to evaluate vec2[[u[1]]] too early by using the function Indexed[vec2,u[1]] . The working MWE would then look like the following: vars = u@# & /@ Range[3]; cons = Flatten@{ Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; NMinimize[ {Total@((vec1[[#]] - Indexed[vec2, u[#]])^2 & /@ R...

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...

What is and isn't a valid variable specification for Manipulate?

I have an expression whose terms have arguments (representing subscripts), like this: myExpr = A[0] + V[1,T] I would like to put it inside a Manipulate to see its value as I move around the parameters. (The goal is eventually to plot it wrt one of the variables inside.) However, Mathematica complains when I set V[1,T] as a manipulated variable: Manipulate[Evaluate[myExpr], {A[0], 0, 1}, {V[1, T], 0, 1}] (*Manipulate::vsform: Manipulate argument {V[1,T],0,1} does not have the correct form for a variable specification. >> *) As a workaround, if I get rid of the symbol T inside the argument, it works fine: Manipulate[ Evaluate[myExpr /. T -> 15], {A[0], 0, 1}, {V[1, 15], 0, 1}] Why this behavior? Can anyone point me to the documentation that says what counts as a valid variable? And is there a way to get Manpiulate to accept an expression with a symbolic argument as a variable? Investigations I've done so far: I tried using variableQ from this answer , but it says V[1...