Skip to main content

string manipulation - Why the pattern match don't work as expected


When I try to figure out this problem,I write such code:


words = Catenate[
WordList[#, Language -> "English",
IncludeInflections -> True] & /@ {"KnownWords", "Stopwords"}];
string = StringReplace[
"What is the best approach to a problem like this in Mathematica?",
" " -> ""];
StringCases[string, __?(MemberQ[words, ToLowerCase[#]] &)]



{"WhatisthebestapproachtoaproblemlikethisinMathematica"}



But I am confused why I get whole string?What's mistake I have made?



Answer



You commented on July 6th:



but I don't understand still why the ? cannot work totally and give whole string.




As MarcoB already quoted:



In a form such as __?test, every element in the sequence matched by __ must yield True when test is applied.



You can easily see for yourself that this is true.


words = {"is", "a", "problem"};

StringCases["What is the best approach to a problem?", __?(MemberQ[words, #] &)]



{"a", "a", "a", "a"}

More explicitly we can use Print or Sow as the test function(1) to see exactly which expressions are being tested:


Reap[ StringCases["Mathematica", __?Sow] ][[2, 1]]


{"M", "M", "M", "M", "M", "M", "M", "M", "M", "M", "M", "a", "a", "a", "a", "a", 
"a", "a", "a", "a", "a", "t", "t", "t", "t", "t", "t", "t", "t", "t", "h", "h",
"h", "h", "h", "h", "h", "h", "e", "e", "e", "e", "e", "e", "e", "m", "m", "m",
"m", "m", "m", "a", "a", "a", "a", "a", "t", "t", "t", "t", "i", "i", "i", "c",

"c", "a"}

Observe that:



  1. Only single letter strings are ever tested

  2. 66 matches are attempted due to every test failing (11 + 10 + 9 + 8 ...)


The first point is actually very useful behavior and I direct you to my own answer Using a PatternTest versus a Condition for pattern matching for additional examples.


The second point is the deleterious consequence of extremely flexible pattern matching used in Mathematica which allows the test function itself to be stateful. I personally feel that there should be a more efficient matching scheme available as an alternative as many uses do not require this level of generality.


Contrast this with Condition (short form /;)



Reap[ StringCases["Mathematica", x__ /; Sow[x]] ][[2, 1]]

{"Mathematica", "Mathematic", "Mathemati", "Mathemat", "Mathema", "Mathem",
"Mathe", "Math", "Mat", "Ma", "M", "athematica", "athematic", "athemati",
"athemat", "athema", "athem", "athe", "ath", "at", "a", "thematica", "thematic",
"themati", "themat", "thema", "them", "the", "th", "t", "hematica", "hematic",
"hemati", "hemat", "hema", "hem", "he", "h", "ematica", "ematic", "emati", "emat",
"ema", "em", "e", "matica", "matic", "mati", "mat", "ma", "m", "atica", "atic",
"ati", "at", "a", "tica", "tic", "ti", "t", "ica", "ic", "i", "ca", "c", "a"}


Here we see that every possible alignment is tried, with the entire candidate sequence passed to the test function each time.


Comments

Popular posts from this blog

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]

plotting - Filling between two spheres in SphericalPlot3D

Manipulate[ SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, Mesh -> None, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], {n, 0, 1}] I cant' seem to be able to make a filling between two spheres. I've already tried the obvious Filling -> {1 -> {2}} but Mathematica doesn't seem to like that option. Is there any easy way around this or ... Answer There is no built-in filling in SphericalPlot3D . One option is to use ParametricPlot3D to draw the surfaces between the two shells: Manipulate[ Show[SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], ParametricPlot3D[{ r {Sin[t] Cos[1.5 Pi], Sin[t] Sin[1.5 Pi], Cos[t]}, r {Sin[t] Cos[0 Pi], Sin[t] Sin[0 Pi], Cos[t]}}, {r, 1, 2 - n}, {t, 0, Pi}, PlotStyle -> Yellow, Mesh -> {2, 15}]], {n, 0, 1}]

plotting - Mathematica: 3D plot based on combined 2D graphs

I have several sigmoidal fits to 3 different datasets, with mean fit predictions plus the 95% confidence limits (not symmetrical around the mean) and the actual data. I would now like to show these different 2D plots projected in 3D as in but then using proper perspective. In the link here they give some solutions to combine the plots using isometric perspective, but I would like to use proper 3 point perspective. Any thoughts? Also any way to show the mean points per time point for each series plus or minus the standard error on the mean would be cool too, either using points+vertical bars, or using spheres plus tubes. Below are some test data and the fit function I am using. Note that I am working on a logit(proportion) scale and that the final vertical scale is Log10(percentage). (* some test data *) data = Table[Null, {i, 4}]; data[[1]] = {{1, -5.8}, {2, -5.4}, {3, -0.8}, {4, -0.2}, {5, 4.6}, {1, -6.4}, {2, -5.6}, {3, -0.7}, {4, 0.04}, {5, 1.0}, {1, -6.8}, {2, -4.7}, {3, -1....