Question

我有来自netCDF文件的纬度LAT，经度LON和windspeed。

我想在给定的LAT，LON坐标%Location of Met Mast 51.94341,1.922094888找到风速。我想分别在RefLAT和RefLON矩阵中找到与LAT和LON最接近的值。当我在LAT和LON中得到最接近的值时，我将使用地址在这个位置找到我的风速。

当我使用下面的代码时，我希望每个列和行值CLON, CLAT, RLAT和RLON都有一个值。相反，我将CLON, CLAT, RLAT和RLON作为972个值的数组，我搜索的矩阵LAT和LON的大小为848 x 972.

LAT = ncread('wind_level2.nc','latitude');
LON = ncread('wind_level2.nc','longitude');
wind = ncread('wind_level2.nc','wind');

LAT = double(LAT);
LON =double(LON);
%Location of Met Mast 51.94341,1.922094888

RefLAT=51.94341;
LATcalc = abs(LAT - RefLAT);
[RLAT,CLAT]=find(min(LATcalc));

RefLON=1.922094888;
LONcalc = abs(LON - RefLON);
[RLON,CLON]=find(min(LONcalc));`

任何帮助表示赞赏。感谢

按要求提供数据样本：

LAT: 848x972 double:

51.6652641296387    51.6608505249023    51.6564369201660    51.6520233154297    51.6476097106934    51.6431961059570    51.6387825012207
    51.6663322448731    51.6619186401367    51.6575050354004    51.6530914306641    51.6486778259277    51.6442642211914    51.6398506164551
    51.6674041748047    51.6629867553711    51.6585731506348    51.6541595458984    51.6497459411621    51.6453323364258    51.6409187316895
    51.6684722900391    51.6640548706055    51.6596412658691    51.6552276611328    51.6508140563965    51.6464004516602    51.6419868469238
    51.6695404052734    51.6651229858398    51.6607093811035    51.6562957763672    51.6518821716309    51.6474685668945    51.6430549621582
    51.6706047058106    51.6661911010742    51.6617774963379    51.6573638916016    51.6529502868652    51.6485366821289    51.6441192626953
    51.6716728210449    51.6672592163086    51.6628456115723    51.6584320068359    51.6540145874023    51.6496009826660    51.6451873779297
    51.6727409362793    51.6683235168457    51.6639099121094    51.6594963073731    51.6550827026367    51.6506690979004    51.6462554931641
    51.6738052368164    51.6693916320801    51.6649780273438    51.6605644226074    51.6561470031738    51.6517333984375    51.6473197937012
    51.6748695373535    51.6704559326172    51.6660423278809    51.6616287231445    51.6572151184082    51.6528015136719    51.6483840942383
    51.6759376525879    51.6715240478516    51.6671066284180    51.6626930236816    51.6582794189453    51.6538658142090    51.6494522094727

LON 848x972 double:

3.04663085937500    3.04491543769836    3.04320049285889    3.04148554801941    3.03977084159851    3.03805613517761    3.03634166717529
3.03959774971008    3.03788304328918    3.03616857528687    3.03445434570313    3.03274011611938    3.03102612495422    3.02931237220764
3.03256440162659    3.03085041046143    3.02913665771484    3.02742290496826    3.02570939064026    3.02399611473084    3.02228283882141
3.02553081512451    3.02381753921509    3.02210426330566    3.02039122581482    3.01867818832397    3.01696562767029    3.01525306701660
3.01849699020386    3.01678419113159    3.01507163047791    3.01335930824280    3.01164698600769    3.00993490219116    3.00822281837463
3.01146292686462    3.00975084304810    3.00803875923157    3.00632715225220    3.00461530685425    3.00290393829346    3.00119256973267
3.00442862510681    3.00271701812744    3.00100588798523    2.99929451942444    2.99758362770081    2.99587273597717    2.99416208267212
2.99739408493042    2.99568319320679    2.99397253990173    2.99226188659668    2.99055147171021    2.98884129524231    2.98713135719299
2.99035930633545    2.98864889144897    2.98693895339966    2.98522901535034    2.98351931571960    2.98180961608887    2.98010015487671
2.98332428932190    2.98161458969116    2.97990512847900    2.97819590568543    2.97648668289185    2.97477769851685    2.97306895256043

Answer 1

如@excaza所述，find的语法在您的情况下不合适。如果没有任何比较运算符，find将采用逻辑比较，并返回包含所有非零元素索引的向量。那不是你想要的。

尝试以下方法：

[RLAT,CLAT]=find(LATcalc<eps);

其中eps是容错，例如0.00001。

你不能给find一个完美的平等，因为你正在寻找的数字可能不存在于矩阵中。使用abs(LAT - RefLAT)，您将拥有两个值之间的差异矩阵。和以前一样，你可能没有一个完美的零，所以你可以找到最接近零的误差容差，以确保捕获最小值。

MATLAB查找最接近指定值的行和列索引

1 个答案: