使用 nco 将 NetCDF 中的经度从 0:360 转换为 -180:180
Converting longitude in NetCDF from 0:360 to -180:180 using nco
我有两个 NetCDF 文件，每个文件代表不同的年份，我想将它们连接起来。它们似乎具有相同的结构，以时间为无限维度，只是其中一个的经度在 0:360 的范围内，而另一个的范围在 -180:180 的范围内。
This question 解决了如何使用 R 中的栅格包处理此问题，但我想使用 nco。
我找到了转换 -180:180-->0:360 的说明（在下面提供），但我对语法不够熟悉，无法让我在另一个方向上反转指令。
关于我的 NetCDF 文件的详细信息：
netcdf soda3.3.1_1980_2015_sst {
dimensions:
depth = 1 ;
latitude = 330 ;
longitude = 720 ;
time = UNLIMITED ; // (432 currently)
variables:
float depth(depth) ;
depth:long_name = "depth" ;
depth:units = "m" ;
depth:axis = "Z" ;
float latitude(latitude) ;
latitude:long_name = "latitude" ;
latitude:units = "degrees_north" ;
latitude:axis = "Y" ;
float longitude(longitude) ;
longitude:long_name = "longitude" ;
longitude:units = "degrees_east" ;
longitude:modulo = 360.f ;
longitude:axis = "X" ;
float temp(time, depth, latitude, longitude) ;
temp:long_name = "Potential temperature" ;
temp:units = "degrees C" ;
temp:valid_range = -10.f, 500.f ;
temp:missing_value = -1.e+20f ;
temp:cell_methods = "time: mean" ;
temp:standard_name = "sea_water_potential_temperature" ;
float time(time) ;
time:long_nime = "time" ;
time:units = "month" ;
data:
longitude = 0.25, 0.75, 1.25, 1.75, 2.25, 2.75, 3.25, 3.75, 4.25, 4.75, 5.25, 5.75, 6.25, 6.75, 7.25, 7.75, 8.25, 8.75, 9.25, 9.75, 10.25, 10.75, 11.25, 11.75, 12.25, 12.75, 13.25, 13.75, 14.25, 14.75, 15.25, 15.75, 16.25, 16.75, 17.25, 17.75, 18.25, 18.75, 19.25, 19.75, 20.25, 20.75, 21.25, 21.75, 22.25, 22.75, 23.25, 23.75, 24.25, 24.75, 25.25, 25.75, 26.25, 26.75, 27.25, 27.75, 28.25, 28.75, 29.25, 29.75, 30.25, 30.75, 31.25, 31.75, 32.25, 32.75, 33.25, 33.75, 34.25, 34.75, 35.25, 35.75, 36.25, 36.75, 37.25, 37.75, 38.25, 38.75, 39.25, 39.75, 40.25, 40.75, 41.25, 41.75, 42.25, 42.75, 43.25, 43.75, 44.25, 44.75, 45.25, 45.75, 46.25, 46.75, 47.25, 47.75, 48.25, 48.75, 49.25, 49.75, 50.25, 50.75, 51.25, 51.75, 52.25, 52.75, 53.25, 53.75, 54.25, 54.75, 55.25, 55.75, 56.25, 56.75, 57.25, 57.75, 58.25, 58.75, 59.25, 59.75, 60.25, 60.75, 61.25, 61.75, 62.25, 62.75, 63.25, 63.75, 64.25, 64.75, 65.25, 65.75, 66.25, 66.75, 67.25, 67.75, 68.25, 68.75, 69.25, 69.75, 70.25, 70.75, 71.25, 71.75, 72.25, 72.75, 73.25, 73.75, 74.25, 74.75, 75.25, 75.75, 76.25, 76.75, 77.25, 77.75, 78.25, 78.75, 79.25, 79.75, 80.25, 80.75, 81.25, 81.75, 82.25, 82.75, 83.25, 83.75, 84.25, 84.75, 85.25, 85.75, 86.25, 86.75, 87.25, 87.75, 88.25, 88.75, 89.25, 89.75, 90.25, 90.75, 91.25, 91.75, 92.25, 92.75, 93.25, 93.75, 94.25, 94.75, 95.25, 95.75, 96.25, 96.75, 97.25, 97.75, 98.25, 98.75, 99.25, 99.75, 100.25, 100.75, 101.25, 101.75, 102.25, 102.75, 103.25, 103.75, 104.25, 104.75, 105.25, 105.75, 106.25, 106.75, 107.25, 107.75, 108.25, 108.75, 109.25, 109.75, 110.25, 110.75, 111.25, 111.75, 112.25, 112.75, 113.25, 113.75, 114.25, 114.75, 115.25, 115.75, 116.25, 116.75, 117.25, 117.75, 118.25, 118.75, 119.25, 119.75, 120.25, 120.75, 121.25, 121.75, 122.25, 122.75, 123.25, 123.75, 124.25, 124.75, 125.25, 125.75, 126.25, 126.75, 127.25, 127.75, 128.25, 128.75, 129.25, 129.75, 130.25, 130.75, 131.25, 131.75, 132.25, 132.75, 133.25, 133.75, 134.25, 134.75, 135.25, 135.75, 136.25, 136.75, 137.25, 137.75, 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200.75, 201.25, 201.75, 202.25, 202.75, 203.25, 203.75, 204.25, 204.75, 205.25, 205.75, 206.25, 206.75, 207.25, 207.75, 208.25, 208.75, 209.25, 209.75, 210.25, 210.75, 211.25, 211.75, 212.25, 212.75, 213.25, 213.75, 214.25, 214.75, 215.25, 215.75, 216.25, 216.75, 217.25, 217.75, 218.25, 218.75, 219.25, 219.75, 220.25, 220.75, 221.25, 221.75, 222.25, 222.75, 223.25, 223.75, 224.25, 224.75, 225.25, 225.75, 226.25, 226.75, 227.25, 227.75, 228.25, 228.75, 229.25, 229.75, 230.25, 230.75, 231.25, 231.75, 232.25, 232.75, 233.25, 233.75, 234.25, 234.75, 235.25, 235.75, 236.25, 236.75, 237.25, 237.75, 238.25, 238.75, 239.25, 239.75, 240.25, 240.75, 241.25, 241.75, 242.25, 242.75, 243.25, 243.75, 244.25, 244.75, 245.25, 245.75, 246.25, 246.75, 247.25, 247.75, 248.25, 248.75, 249.25, 249.75, 250.25, 250.75, 251.25, 251.75, 252.25, 252.75, 253.25, 253.75, 254.25, 254.75, 255.25, 255.75, 256.25, 256.75, 257.25, 257.75, 258.25, 258.75, 259.25, 259.75, 260.25, 260.75, 261.25, 261.75, 262.25, 262.75, 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325.75, 326.25, 326.75, 327.25, 327.75, 328.25, 328.75, 329.25, 329.75, 330.25, 330.75, 331.25, 331.75, 332.25, 332.75, 333.25, 333.75, 334.25, 334.75, 335.25, 335.75, 336.25, 336.75, 337.25, 337.75, 338.25, 338.75, 339.25, 339.75, 340.25, 340.75, 341.25, 341.75, 342.25, 342.75, 343.25, 343.75, 344.25, 344.75, 345.25, 345.75, 346.25, 346.75, 347.25, 347.75, 348.25, 348.75, 349.25, 349.75, 350.25, 350.75, 351.25, 351.75, 352.25, 352.75, 353.25, 353.75, 354.25, 354.75, 355.25, 355.75, 356.25, 356.75, 357.25, 357.75, 358.25, 358.75, 359.25, 359.75 ;
这是我在手册中找到的将 -180:180 转换为 0:360 的代码。
在此背景下，可以看到以下命令足以将输入文件旋转 180 度经度：
% ncks -O -v LatLon --msa -d Lon,0.,180. -d Lon,-180.,-1.0 in.nc out.nc
% ncap2 -O -s 'where(Lon < 0) Lon=Lon+360' out.nc out.nc
% ncks --trd -C -H -v LatLon ~/nco/data/in.nc
Lat[0]=-45 Lon[0]=-180 LatLon[0]=0
Lat[0]=-45 Lon[1]=-90 LatLon[1]=1
Lat[0]=-45 Lon[2]=0 LatLon[2]=2
Lat[0]=-45 Lon[3]=90 LatLon[3]=3
Lat[1]=45 Lon[0]=-180 LatLon[4]=4
Lat[1]=45 Lon[1]=-90 LatLon[5]=5
Lat[1]=45 Lon[2]=0 LatLon[6]=6
Lat[1]=45 Lon[3]=90 LatLon[7]=7
% ncks --trd -C -H -v LatLon ~/out.nc
Lat[0]=-45 Lon[0]=0 LatLon[0]=2
Lat[0]=-45 Lon[1]=90 LatLon[1]=3
Lat[0]=-45 Lon[2]=180 LatLon[2]=0
Lat[0]=-45 Lon[3]=270 LatLon[3]=1
Lat[1]=45 Lon[0]=0 LatLon[4]=6
Lat[1]=45 Lon[1]=90 LatLon[5]=7
Lat[1]=45 Lon[2]=180 LatLon[6]=4
Lat[1]=45 Lon[3]=270 LatLon[7]=5
提前致谢！
如果字段是全局的，您可以使用
从 0-360 转换为 -180,180
cdo sellonlatbox,-180,180,-90,90 in.nc out.nc
并转换为另一种方式：
cdo sellonlatbox,0,360,-90,90 in.nc out.nc
不涉及插值，应该只是改变lon坐标。
好问题。尝试
ncks -O --msa -d lon,181.,360. -d lon,0.,180.0 in.nc out.nc
ncap2 -O -s 'where(lon > 180) lon=lon-360' out.nc out.nc
第一条命令移动数据，第二条命令重新校准坐标到新移动的数据。关于应用此算法的评论：
小心指定不重叠的半球，例如，通过在第一个命令中无意中指定坐标范围，这两个命令都包含日期线。
一些用户会发现使用基于索引而不是基于坐标的 hyperslabs 使这一点更加清晰。检查田地图以确保您的旋转是正确的。
我通过将一个数据的网格设置为另一个来解决了 CDO 中的类似问题。例如，如果您希望两个文件具有相同的 -180:180 表示法，那么试试这个：
cdo setgrid,file_with_-180:180_grid.nc file_with_0-360_grid.nc new_file.nc
我有两个 NetCDF 文件，每个文件代表不同的年份，我想将它们连接起来。它们似乎具有相同的结构，以时间为无限维度，只是其中一个的经度在 0:360 的范围内，而另一个的范围在 -180:180 的范围内。
This question 解决了如何使用 R 中的栅格包处理此问题，但我想使用 nco。
我找到了转换 -180:180-->0:360 的说明（在下面提供），但我对语法不够熟悉，无法让我在另一个方向上反转指令。
关于我的 NetCDF 文件的详细信息：
netcdf soda3.3.1_1980_2015_sst {
dimensions:
depth = 1 ;
latitude = 330 ;
longitude = 720 ;
time = UNLIMITED ; // (432 currently)
variables:
float depth(depth) ;
depth:long_name = "depth" ;
depth:units = "m" ;
depth:axis = "Z" ;
float latitude(latitude) ;
latitude:long_name = "latitude" ;
latitude:units = "degrees_north" ;
latitude:axis = "Y" ;
float longitude(longitude) ;
longitude:long_name = "longitude" ;
longitude:units = "degrees_east" ;
longitude:modulo = 360.f ;
longitude:axis = "X" ;
float temp(time, depth, latitude, longitude) ;
temp:long_name = "Potential temperature" ;
temp:units = "degrees C" ;
temp:valid_range = -10.f, 500.f ;
temp:missing_value = -1.e+20f ;
temp:cell_methods = "time: mean" ;
temp:standard_name = "sea_water_potential_temperature" ;
float time(time) ;
time:long_nime = "time" ;
time:units = "month" ;
data:
longitude = 0.25, 0.75, 1.25, 1.75, 2.25, 2.75, 3.25, 3.75, 4.25, 4.75, 5.25, 5.75, 6.25, 6.75, 7.25, 7.75, 8.25, 8.75, 9.25, 9.75, 10.25, 10.75, 11.25, 11.75, 12.25, 12.75, 13.25, 13.75, 14.25, 14.75, 15.25, 15.75, 16.25, 16.75, 17.25, 17.75, 18.25, 18.75, 19.25, 19.75, 20.25, 20.75, 21.25, 21.75, 22.25, 22.75, 23.25, 23.75, 24.25, 24.75, 25.25, 25.75, 26.25, 26.75, 27.25, 27.75, 28.25, 28.75, 29.25, 29.75, 30.25, 30.75, 31.25, 31.75, 32.25, 32.75, 33.25, 33.75, 34.25, 34.75, 35.25, 35.75, 36.25, 36.75, 37.25, 37.75, 38.25, 38.75, 39.25, 39.75, 40.25, 40.75, 41.25, 41.75, 42.25, 42.75, 43.25, 43.75, 44.25, 44.75, 45.25, 45.75, 46.25, 46.75, 47.25, 47.75, 48.25, 48.75, 49.25, 49.75, 50.25, 50.75, 51.25, 51.75, 52.25, 52.75, 53.25, 53.75, 54.25, 54.75, 55.25, 55.75, 56.25, 56.75, 57.25, 57.75, 58.25, 58.75, 59.25, 59.75, 60.25, 60.75, 61.25, 61.75, 62.25, 62.75, 63.25, 63.75, 64.25, 64.75, 65.25, 65.75, 66.25, 66.75, 67.25, 67.75, 68.25, 68.75, 69.25, 69.75, 70.25, 70.75, 71.25, 71.75, 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325.75, 326.25, 326.75, 327.25, 327.75, 328.25, 328.75, 329.25, 329.75, 330.25, 330.75, 331.25, 331.75, 332.25, 332.75, 333.25, 333.75, 334.25, 334.75, 335.25, 335.75, 336.25, 336.75, 337.25, 337.75, 338.25, 338.75, 339.25, 339.75, 340.25, 340.75, 341.25, 341.75, 342.25, 342.75, 343.25, 343.75, 344.25, 344.75, 345.25, 345.75, 346.25, 346.75, 347.25, 347.75, 348.25, 348.75, 349.25, 349.75, 350.25, 350.75, 351.25, 351.75, 352.25, 352.75, 353.25, 353.75, 354.25, 354.75, 355.25, 355.75, 356.25, 356.75, 357.25, 357.75, 358.25, 358.75, 359.25, 359.75 ;
这是我在手册中找到的将 -180:180 转换为 0:360 的代码。
在此背景下，可以看到以下命令足以将输入文件旋转 180 度经度：
% ncks -O -v LatLon --msa -d Lon,0.,180. -d Lon,-180.,-1.0 in.nc out.nc
% ncap2 -O -s 'where(Lon < 0) Lon=Lon+360' out.nc out.nc
% ncks --trd -C -H -v LatLon ~/nco/data/in.nc
Lat[0]=-45 Lon[0]=-180 LatLon[0]=0
Lat[0]=-45 Lon[1]=-90 LatLon[1]=1
Lat[0]=-45 Lon[2]=0 LatLon[2]=2
Lat[0]=-45 Lon[3]=90 LatLon[3]=3
Lat[1]=45 Lon[0]=-180 LatLon[4]=4
Lat[1]=45 Lon[1]=-90 LatLon[5]=5
Lat[1]=45 Lon[2]=0 LatLon[6]=6
Lat[1]=45 Lon[3]=90 LatLon[7]=7
% ncks --trd -C -H -v LatLon ~/out.nc
Lat[0]=-45 Lon[0]=0 LatLon[0]=2
Lat[0]=-45 Lon[1]=90 LatLon[1]=3
Lat[0]=-45 Lon[2]=180 LatLon[2]=0
Lat[0]=-45 Lon[3]=270 LatLon[3]=1
Lat[1]=45 Lon[0]=0 LatLon[4]=6
Lat[1]=45 Lon[1]=90 LatLon[5]=7
Lat[1]=45 Lon[2]=180 LatLon[6]=4
Lat[1]=45 Lon[3]=270 LatLon[7]=5
提前致谢！
如果字段是全局的，您可以使用
从 0-360 转换为 -180,180cdo sellonlatbox,-180,180,-90,90 in.nc out.nc
并转换为另一种方式：
cdo sellonlatbox,0,360,-90,90 in.nc out.nc
不涉及插值，应该只是改变lon坐标。
好问题。尝试
ncks -O --msa -d lon,181.,360. -d lon,0.,180.0 in.nc out.nc
ncap2 -O -s 'where(lon > 180) lon=lon-360' out.nc out.nc
第一条命令移动数据，第二条命令重新校准坐标到新移动的数据。关于应用此算法的评论： 小心指定不重叠的半球，例如，通过在第一个命令中无意中指定坐标范围，这两个命令都包含日期线。 一些用户会发现使用基于索引而不是基于坐标的 hyperslabs 使这一点更加清晰。检查田地图以确保您的旋转是正确的。
我通过将一个数据的网格设置为另一个来解决了 CDO 中的类似问题。例如，如果您希望两个文件具有相同的 -180:180 表示法，那么试试这个：
cdo setgrid,file_with_-180:180_grid.nc file_with_0-360_grid.nc new_file.nc