MATLAB 查找最接近指定值的行和列索引
MATLAB find row and column index of closest to specified value
我有来自 netCDF 文件的纬度 LAT
、经度 LON
和风速。
我想找到给定 LAT、LON 坐标 %Location of Met Mast 51.94341,1.922094888
处的风速。我试图分别在 LAT
和 LON
矩阵中找到最接近 RefLAT
和 RefLON
的值。当我在 LAT 和 LON 中获得最接近的值时,我将使用这些地址来定位我在该位置的风速。
当我使用下面的代码时,我希望每个列值和行值 CLON, CLAT, RLAT
和 RLON
都有一个值。相反,我得到 CLON, CLAT, RLAT
和 RLON
作为 972 个值的数组,我正在搜索的矩阵 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));`
感谢任何帮助。谢谢
请求的数据样本:
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
正如@excaza 所指出的,您的 find
语法不适合您的情况。没有任何比较运算符,find
假定逻辑比较并将 return 包含所有非零元素索引的向量。那不是你想要的。
尝试以下操作:
[RLAT,CLAT]=find(LATcalc<eps);
其中 eps
是误差容限,例如 0.00001。
您不能给出 find
完全相等,因为您要查找的数字可能不在矩阵中。使用 abs(LAT - RefLAT)
,您将得到两个值之间的差值矩阵。和以前一样,您可能没有完美的零,因此您可以通过足够低的误差容限找到最接近零的结果,以确保达到最小值。
我有来自 netCDF 文件的纬度 LAT
、经度 LON
和风速。
我想找到给定 LAT、LON 坐标 %Location of Met Mast 51.94341,1.922094888
处的风速。我试图分别在 LAT
和 LON
矩阵中找到最接近 RefLAT
和 RefLON
的值。当我在 LAT 和 LON 中获得最接近的值时,我将使用这些地址来定位我在该位置的风速。
当我使用下面的代码时,我希望每个列值和行值 CLON, CLAT, RLAT
和 RLON
都有一个值。相反,我得到 CLON, CLAT, RLAT
和 RLON
作为 972 个值的数组,我正在搜索的矩阵 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));`
感谢任何帮助。谢谢
请求的数据样本:
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
正如@excaza 所指出的,您的 find
语法不适合您的情况。没有任何比较运算符,find
假定逻辑比较并将 return 包含所有非零元素索引的向量。那不是你想要的。
尝试以下操作:
[RLAT,CLAT]=find(LATcalc<eps);
其中 eps
是误差容限,例如 0.00001。
您不能给出 find
完全相等,因为您要查找的数字可能不在矩阵中。使用 abs(LAT - RefLAT)
,您将得到两个值之间的差值矩阵。和以前一样,您可能没有完美的零,因此您可以通过足够低的误差容限找到最接近零的结果,以确保达到最小值。