javascript – 从任意字符串生成图像
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内容图文
我已经看到了很多ascii艺术转换器的图像,但我需要的是更复杂一点.而不是从ascii字符创建的马赛克图像,我需要能够使用任意长度的预定义字符串集作为我的马赛克拼贴.
我已经做了很多谷歌搜索,但我甚至不知道如何构建我的查询？有这样的事吗？如果它不会在其图像生成中使用重复,那么奖励.
解决方法:
不太适合Stack Overflow,但不过是一个有趣的项目.所以我有一点乐趣,看看我能走得多远.
我认为归结为：
>计算一组ASCII字符的灰度值;
>计算每个字符串的“最佳拟合”;
>重复2.直到完成.
我不认为“没有重复”是可行的,除非你有非常小的图像和许多候选字符串.这就是我想出来的,使用我的头像和我当前的徽章作为字符串列表. (我排除了我的“c”徽章;不知怎的,我的程序决定对大补丁来说“最适合”,这不是很有吸引力.精神笔记：不包括单字符串.)
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你必须眯一点;在小尺寸它看起来像这样：
这是我创建它的方式.
第1步：找到合适的图像;
第2步：转换为灰度;
第3步：将灰色转换为极值.此步骤是为了确保输入范围(灰度值)使用0到255的整个范围.
第4步：调整图像大小以获得最佳效果.我选择了80×40,故意将图像压缩了一半.这是因为“文本”通常高于宽度.不同的字体需要不同的宽高比！ 80将是每行的字符数,40是总行数.
我使用Photoshop进行上述步骤,只是因为我不想为它编写代码.只要您可以访问原始图像数据,这并不困难,但这需要大量的工作而且不是很有趣.
中间步骤,放大400％,以便您可以看到像素：
第5步：找到一个等宽的位图字体.我在网上的某个地方找到了一个不错的8×8.也许更大的尺寸可能更好,但只是非常小,因为限制因素是你的字符串,而不是字体.
步骤6：计算每个ASCII字符的“灰色”值.这是黑色像素的数量除以总像素数.为了更好的传播,我将每个字符的结果除以找到的最大值,因此最低值为0(对于空格),最高值为1(恰好是M,但它取决于字体).然后我将这些值乘以255,因此它模仿灰度值.最后,由于这些值与灰度图像中的值相反,我将其替换为255值.
在这之间,我做了很多测试,以确保我的初步想法仍然是健全的.这是使用简单的灰色到字符的翻译转储我的测试图像：
0MMMMMM00000000000RDRRDDDDDDDDDDDDDDR#@RRRRR#00000RRR@RRRRR@RR@RRRRD&44�#0000
MMMMMM000000000000RDRDDDDDDDDDDDDDDDR0####0R&&&&&&DR@RRR@#########RRDDD4@0000000
MMM000#00000000000RDRRDDDDDDDDDDD&&&R00#000@&DR@####RR@#0000000##0#RD4PP&R@00000
M000#0000000000000RDDDDDDDDDDD&D&&D&R###0000@#0000#@@##@@@@@##000000R&2FPP4R#000
00###0000000000000RDDDDDDDDDDDDDD&&&D0000#@##@##00##@RDD&DRRRR#00000@R4FFP4PD@00
00#000000000000000RDDDDDDDDDD&D&D&&&DRRRRDR@@@##00#R&4&44DDDRRR#0000##R&FPP4PRR@
00##00#00000000000RDDDDDDDDDDD&&&&&&RDRRRRRRRR##@@RR&F4&RDDRR@@#0000000#&44&&4FD
000000000000000000DDDDDDD&&&DD&&&D&&R##@RRRRRRRRRRRRD&DRD&44DD&RR@#00000RDDRR4F4
000000000000000000RDDDDDDDRRDRDDD&DR@###@R&4&4444PP4DRRR&P22FF2FP&DRRRRRRRR@R4FD
000000000000000000RRRRRRR@00@@R&&DRRRRRR@&P2$33*33*4RRRDP23$*33$$**333*$$2D#@4DR
000000000000000000RRRRRRR@0000#DDRRRRRD&D4*$$%%ff3F&DRRR2$1%$$%11ff%ll1l;'IRDF&R
000000000000000000RRR@#0#00000R4444DRRRR&3llf33$32PDRR&3Ii(i%fIlI1Il!ii/' .FF4DR
0000000#0000000000#@@#0MM0M00#RP22*24DR4$l!!!I%*P4DDP31i===/lf1Illi((ii=;..*F4RR
000000000000000000#@@##0M00000@F*$%*2PFflilllI1f3*2PF3fIi/=(lf$fIi/======;iF&P4R
000000000000000000##@##00000000&3f$2F2*1i!ll1$2P4RD&4P*$$%!ii!I!ii/=/=;;;(*&R4FR
000000000000000000#####00000000D%I3$$fflil!l1$3%3&RD2PP*$%1ll!lI1%f$3fI="1RRRRDR
000000000000000000#####00000000Rf1f%II1IIl!l!!1%ff$*FFFF*%((lf**F4&D&PPFf2RRRRRR
00000000000000000M#####0000000003!i11II11l!!!(ilIIl1f$f%I="=1FP4&@#R&DP22&RRR@#@
0000000000000000000###0000000000#$i/iIIIl!!i=;"";===;""""";i12FFP4&4*4&*FRRRR@@R
000000000000000000M000000000000000R4FF1l!i(=;"''"""""'...';/i$3$f$$33$$$4DRRRRRR
000000000000000000000000000M00000000@21l(/ii/=;;=ilI;"'...'=;I%11f$$$ffP@@RR@R#0
000000000000000000000000000M00000000&11!=/li(i!I%%Ii;... .";"!I!!I1ff$*R0#@#0000
000000000000000000000000M0000M000000$lli=//=/i!lIi/l%I/""=//'=Iiiil%f$PR@#@00000
000000000000000000000000M0M0M00000MRl!!!/=;;/l%fl!lI%3233$ff%11II%%f*2PP@##00000
00000000000000000M000000M0MM00000003=illl!!iI%11%%f%f$32FF22$1%ff$f2DD4PR0000000
00000000000000000M00M0#RRRRRRRRRRRPi/(lllI1I(=;=iI$2222FPFFP*$fff%FRR2322&R00000
00000000000000000MM00#RRD&&&&&444*l/(i!ll!i!==/(il%$33*2FFF23$ff*4DDDF33*22&0000
0000000000000000000000@R@@@@@#&f!((/i!!!lI!i=;;/!l1f32*3$$$ff$2&RR@###@##00##000
0000000000000000#RRRR@@4F4R0R3!////(iii!!l%Ii=;"=l%f$333$ff%$4############00000#
0000000000000000000000#@@@#0Di/ii(=/(i!!!lI%%1!(iI%$**$$fffF@0000000000000000000
00000000000000000000000000000R3l((/((ii!lll!I%$3$f$$$$$3$f20M0000000000000000000
000000000000000000000000000000#2l///((!!l!!i!!1%f%f$3$f%%fR000000000000000000000
00000000000000#00000000000000000&l//(i!llllII1%%%1I1ff1I1&000000000000000000#00#
M00000000000000000000000000000000R3l!!llII111%%%%%%f$$f%2000000000000000000RR00#
0000#000000000000000000000000000000R*%1II111%%ffff$33fffR000000000000000000##00#
0####000000000000000000000000000000M0RF3ff$$$fff$$33$ff&00000000000#000000##000R
#00000000#00000000000000000000000000000R4FFFF22*****3fF00000000000000000000000@&
00###000000000000000000000000000000000000@RRDD&&&4P*34#00000000000000000000000#R
00000####000000000000000000000000000000000000##@RRDDR00000000000000M0000000000##
00000000#000000000000000000000000000000000000000000000000000000000000000000000##
这类似于您的平均图像到ASCII艺术转换器输出.
第7步：找到自己要使用的字符串列表.
步骤8：从左上角开始,根据图像测试每个字符串的覆盖范围.打印最合适的,通过这个字符串的长度增加位置,重复直到完成. (如果您不想重复,则此时从池中删除该字符串.)
第9步：获利！
对于覆盖测试,我使用(source-dest)²的总和(每个字符/像素)除以字符串的长度：lower = better – 字符串和目标之间的最小差异.
我在这里没有考虑漂亮的行结尾.如果一个单词完全填满一行,我尝试给出一个负面奖励,但输出的差异很小.它仍然可以使用更大的字符串集.
一个可能的改进是测试一系列字符串,即,而不是这里的贪婪方法,使用动态编程方法,就像Donald Knuth设计用来决定自动换行文本中的最佳中断一样.
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