m*l
发帖数: 507 | 1 You can use GoogleEarth.
In Google Earth you can draw point, line, polygon
You can place labels and using styles.
You can export image from google earth and edit the image using other tools
such as Powerpoint.
And it is FREEEEE |
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A*****r
发帖数: 795 | 3 use the raster calculator in Spatial Analysis module.
google it |
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n*****r
发帖数: 2447 | 4 haha, thank you! I finally figure out how to do it. Still runing GeoHMS.... |
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m******t
发帖数: 147 | 6
to
---------------------------
who told you this? did you try gridclip in grid module?
DEM |
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G*******s
发帖数: 76 | 7 孩子们哪里知道命令行，都是用鼠标点
module? |
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C*********g
发帖数: 3728 | 8 that's nice. When I learned AML years ago, my teacher also told me that's
not possible to clip to a shapefile...
module? |
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S****t
发帖数: 1186 | 9 one needs to convert the shapefile to coverage first to use the command
mentioned by myaspect. |
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i****y
发帖数: 12 | 10 Arctoolbox->Spatial analysis->Extraction->Extract by mask |
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G*******s
发帖数: 76 | 11 and you may try overlay your vegetation polygons
over your rainfall in ArcScene to get an
interactive 3D rainfall map. |
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r**a
发帖数: 761 | 12 在arcmap里，怎么作出一个polygon给定距离的inner ring?
我知道buffer可以做出outer ring,然后我试着给了一个负值的距离，可是不行
请大家帮忙，明天要交了，呜呜
实在不行就手画了，可是我不知道直角的地方inner ring的形状是什么？
数学太烂了。。
多谢多谢！ |
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t********r
发帖数: 4908 | 13 wow, mapinfo is still in market? surprised to know........ |
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w*m
发帖数: 1806 | 14 hi，我以前也被这个问题折磨过。帮我导师计算抽水井到最近的河流的最短距离。我在
arcgis里没有找到这样的工具啊。 |
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W**********i
发帖数: 136 | 16 AO里面IProximityOperator.returnDistance可以返回两个Geometry之间的最短距离 |
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W**********i
发帖数: 136 | 17
it
呵呵，你这个方法貌似独立于软件之外了，挺好的啊
我那个就依赖AO帮忙计算了 |
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w*m
发帖数: 1806 | 18 看来你比较有AO经验啊。你是不是做GIS programming的？
对ArcIMS熟不熟啊？ |
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W**********i
发帖数: 136 | 19 对AO比较熟，ArcIMS了解一些
目前正在努力找工作，5555555555555 |
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a*******8
发帖数: 32 | 20 除了上面提到的Voronoi方法外，用ArcGIS对两个图层进行spatial join也可解决此问
题。离多边形最近的点的属性将被赋予该多边形，同时将产生新的field存储点到多边
形的距离。 |
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t********h
发帖数: 161 | 21 我现在有个polygon file，想做成3D的，symbology反映一个attribute，3D高度反映另
一个attribute. 我在arcscene里面尝试了extrusion，效果不理想，surface不smooth
，边界变化比较突兀，所以我想用raster，不过问题是，raster的话，我就不能用
symbology了。不知道各位高手有没有办法解决啊？google，百度了半天了，还是没有
结果。谢过 |
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c*********d
发帖数: 218 | 22 我现在的问题是：怎样得到the shortest distance between a point(crime location
) and a polygon (park)? |
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s*******1
发帖数: 146 | 23 find the centroid of a polygon, then combine with the point data, create a
point shapefile, import the shapefile into Geoda, get spatial weight matrix,
you find the distance between two points, then find the shortest one.
location |
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l*******r
发帖数: 24 | 24 没有用过arcgis里面的这个功能
但是练习编程时实现过
应该就是给一个点集合
生成一个多边形集合，有拓扑或者没有 |
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f******k
发帖数: 297 | 26 来自主题: Mathematics版 - 一个几何题 but m=4 is trivial. shouldn't the number of vertices of the
polygon be in the order of O(m)? |
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e****o
发帖数: 165 | 27 发信人: littless (littless), 信区: Science
标 题: some tales of mathematicans(51)
发信站: 水木社区 (Tue May 23 14:02:57 2006), 站内
尤太民族被认为是世界上最聪明的民族
特别是学术研究和经商方面
所以世界上很多地方都有排斥尤太人的倾向
在苏联尤其如此
下面是莫斯科大学1978年对非尤太人和尤太人
的入学考试试题：
1）write the law of cosines
2）is the function x-|x|^3differentiable at the origin？
3）draw the graph of y=2|x|^(-1)
1)which of the following two number is larger,(413)^(1/3)
or 6+3^(1/3)
2)find the integer solutins to the equations x^y=y^x
3)prove that a convex polygon of surface area equal to |
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s**p
发帖数: 207 | 28 不是拓扑学学问题
就是我有一个mesh,我想做一个sanity check,保证这个mesh是连通的。
这个breadth first search好像complexity 很高，我想最好就是O(N)或者O(1)的
complexity.
如果这个net是在一个平面上，我好像可以做出来，O(1), 但是很可能也错了
V+(F+1)-E=2, 如果成立，就是连通的，如果不，就不是联通。
其实就是一个变形的euler poincare。我加上一个虚拟的polygon把边界给连起来，把
这个2维平面变成一个genus=0的三维封闭体.
不过如果这个区域不是单联通的，似乎这个就不对了。
anyway,没学过图论拓扑的东西，我是做boundary element的
问题现在是一个三维的surface mesh. anyway, if it costs too much, it does not
worth it. i will let the code run until it crashs :( |
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O*******d
发帖数: 20343 | 29 google "point inside polygon" |
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z****f
发帖数: 484 | 30 球体就不能由有限个extreme points线性表示，Krein-Milman theorem也没有这样说。
不过polygon只需要有限个。 |
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w****a
发帖数: 4 | 31 consider the universe of all polygons with three or four sides, and define
the following open statements for this universe.
h(x): all sides of x are equal
p(x): x has as interior angle that exceeds 180
之类，，，
要怎么写？
谢谢！ |
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m****m
发帖数: 2211 | 32 水木高人SayMyName提供的链接：
http://www.springerlink.com/content/e4823102v1914787/
马在他的文章里这么说的
Referring to the author's thesis at Columbia University around 1996, Phong a
nd Stein [56] were able to announce a general conclusion in 1997 that the os
cillatory integral operator in (1) with x,y\in R^1 and real analytic phase f
unction has the same decay rate as Varchenko's conclusion for oscillatory in
tegrals. In the author's thesis, and also in Phong and Stein's paper, the sp
ace is partitioned according ... 阅读全帖 |
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l**x
发帖数: 110 | 33 知道这个版的专业医生很多，所以占宝地求指点。恳请大家一点给点指导，下一步怎么
办合适？
患者44岁，男，6年前因为肝癌做了肝移植 （符合米兰标准），之后无灾无病好几年。
去年2016年初发现轻微贫血，血色素12，因为当时在服小剂量阿司匹林，于是停服，看
看有没好转。托托拉拉到了年底，还是贫血，血色素10.5。医生说做个内镜。于是在胃
内发现一个肿瘤，内镜医生直觉是GIST。随后CT，但CT却没有异常。4月初第二次胃镜
，GIST依然还在，取样活检表明可能肝癌转移或者是非常罕见的HAC。第二次CT 报告大
小46x35mm，说比较第一次32mm变大了。可能第一次看片医生那天喝多了？肿瘤位置在
Fundus，非常接近GOJ。
AFP好久没有查了，因为肝上的情况一直很好，肝移植前AFP 100多，移植后AFP一直小
于2。上一次至少是2，3年前。也是疏忽了。现在AFP 254. 显然胃上这个要么是转移，
要么是HAC。
下周二医生还要安排一次PETCT，周三见外科医生，估计排除转移外就安排切除了。 求
这里的专家教授指点，这个情况如何是好？下面我贴上最近几个报告。
个人感觉肝移植后5，6年转... 阅读全帖 |
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e*******n
发帖数: 4912 | 34 1.在这个系列里我打算写一些我在各种文章和书中看到的八卦
希望能博大家一笑
有一次littlewood问hardy,为什么他每次到一个旅馆就会把镜子用毛巾盖起来?
回答是:因为他长得太丑了
2.Hadamard,Jacques去意大利Bologna开1928年国际数学家大会,期间要坐火车去一个地
方
车厢里有很多人在聊天,他觉得十分累,就出了道困难的数学题,众人思考这道题,
车厢里马上安静下来了,于是Hadamard就可以睡觉了
3.Bourbaki是一个法国数学家的集体代名词
Bourbaki的第一篇文章发表在comptes Rendus(法国科学院的一个杂志)上
在1949年Journal of symbolic logic上的一篇文章
"Foundations of mathematics for the working mathematican"
中,Bourbaki教授的地址是University of Nancago
一个杜撰的地址,分别是Nancy和Chicago(weil在那里)前后组合
1940年,Boas,Ralph(MR的主编)曾经在Encyclopa... 阅读全帖 |
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K****n
发帖数: 5970 | 35 最近临时要帮朋友做一个task,她说最好用matlab
我想问问psycho toolbox或者matlab的基本方程里,用什么画polygon?
多谢大侠鸟~ |
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s***e
发帖数: 267 | 36 不需要这些假定。首先任何直角三角形都是0。不然两个直角三角形可以拼成一个矩形
，矛盾。
然后任何三角形可以分成两个直角三角形，所以也是0。任何polygon可以分成三角形。
F= |
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p*****k
发帖数: 318 | 37
DuGu, i see - sorry that was a naive argument.
here are some thoughts based on what ppl have discussed so far:
since every polygon can be triangulated, and every triangle can be
decomposed into two right triangles, it suffices to prove this
"measure function" vanishes for all right-angled triangles.
now take the middle points of each side, along with the right-angled
vertex. this forms a rectangle; the rest are two similar triangles
of half of the original size. by repeating this, the origina |
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D**u
发帖数: 204 | 38 The question does not require F to be countable additive.
In the traditional measure theory, every measurable set has a non-negative
measure, so if a measurable set S_1 is a subset of S_2, then measure(S_2) >=
measure(S_1). Then for a P in the 2-d plane, we can use a countable union
of rectangles to "approximate" the measure of P.
The same approximation fails in this question because F does not guarantee a
non-negative "measure" on any polygon. |
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D**u
发帖数: 204 | 39 Here is the counterexample (such that F(P) != 0 for some P).
...
Suppose the vertices of polygon P are (in counterclockwise position) z1, ...
, zn, (z_{n+1} := z1). Let theta_i be the angle of vector
z_{i+1} - z_i.
Define g(theta) = sin(3*theta), then we have g(theta) = - g(theta + pi).
We define F(p) = \sum |z_{i+1} - zi| * g(theta_i).
Ease to check the following:
(1) if P is the union of 2 non-intersecting P1 and P2, because the "line integral" on the common boundary of P1 and P2 vanishes, so |
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t*******e
发帖数: 172 | 40 Is not this G can define a measure on the plane?
For any X we can define G(X)= inf \Sigma G(P_{i}), where inf with respcet to
all the coverings of X by polygons.
Maybe we need G is countable summable.
). |
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n******m
发帖数: 169 | 41 not sure.
1. cov(x,y)/var(y)
2. 6/11
3. (99%)^100, estimate 37.5%, calculator says 36.6%.
if corelated, can be max 99% min 0%
4.....
5. Set E(x)=0. Suppose pdf exist. need to prove int(e^xf(x)dx))<=1. Let F(t)
=int(e^(tx)f(x)dx), then F(0)=1, F'(t)<0, So F(1)<=1.
6. Fix one vertex, consider clockwisely the first ray going out from this
vertex, it divids the polygon into 2 part. Then triangulate the 2 parts. To
gaurantee on overcount, the first part should not have any ray coming out
from the... 阅读全帖 |
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l*******1
发帖数: 113 | 43 pls enlighten me about your method, da niu... |
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L**********u
发帖数: 194 | 45 用markov chain 应该很容易做出来。 |
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L**********u
发帖数: 194 | 46 呵呵， 居然还有一篇文章写这个问题。
有用没有，荡了再说。 |
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M*******i
发帖数: 82 | 47 the expected time from state (n-1) to state n is simply n - 1
sum(i-1) i from 1 to n -> n(n-1)/2 |
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h****a
发帖数: 234 | 48 There must be an algorithm for the computer
to distinguish which
triangle is positive and which one is negative
For example if one marks the veritices of the
outer polygon counterclockwise as A1, A2 .. An
and choose some arbitrary point in the plane,
say O
then:
the triangle [O Ai Ai+1] would be positive if
OAi -> OAi+1 goes counterclockwise, and vice
versa.
Thus adding the masses and moments of OAiAi+1
i=1,2,..,n-1 will give the result for the
outside one.
Do the same for the inside one, and su |
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N*********r
发帖数: 40 | 49 First of all, the brightness of a RGB pixel is not simply R+G+B.
( I use capital RGB because they are not normalized )
take YUV for example, Y = 0.299R + 0.587G + 0.114B
Anyway, let's define "brightness" as R+G+B, but the polygon described by
0 <= R <= 255
0 <= B <= 255
0 <= G <= 255
on the plane
R+G+B = c
where c is a constant, is not nessarily a triangle at all. An obvious
(counter) example is c = 3/2 * 255 |
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y****n
发帖数: 2 | 50 I used ESRI free scripts for ArcView GIS before to get polygon centroids,
etc. the site is http://arcscripts.esri.com/. Hope there is something useful
for |
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