Python：将坐标从列表传递到函数
Python: passing coordinates from list to function
我正在使用 workshop 中的一些代码按最接近我指定坐标的坐标从 netCDF 文件中提取数据。当只使用一组坐标时，我可以毫无困难地提取我需要的值，如下所示：
import numpy as np
import netCDF4
from math import pi
from numpy import cos, sin
def tunnel_fast(latvar,lonvar,lat0,lon0):
'''
Find closest point in a set of (lat,lon) points to specified point
latvar - 2D latitude variable from an open netCDF dataset
lonvar - 2D longitude variable from an open netCDF dataset
lat0,lon0 - query point
Returns iy,ix such that the square of the tunnel distance
between (latval[it,ix],lonval[iy,ix]) and (lat0,lon0)
is minimum.
'''
rad_factor = pi/180.0 # for trignometry, need angles in radians
# Read latitude and longitude from file into numpy arrays
latvals = latvar[:] * rad_factor
lonvals = lonvar[:] * rad_factor
ny,nx = latvals.shape
lat0_rad = lat0 * rad_factor
lon0_rad = lon0 * rad_factor
# Compute numpy arrays for all values, no loops
clat,clon = cos(latvals),cos(lonvals)
slat,slon = sin(latvals),sin(lonvals)
delX = cos(lat0_rad)*cos(lon0_rad) - clat*clon
delY = cos(lat0_rad)*sin(lon0_rad) - clat*slon
delZ = sin(lat0_rad) - slat;
dist_sq = delX**2 + delY**2 + delZ**2
minindex_1d = dist_sq.argmin() # 1D index of minimum element
iy_min,ix_min = np.unravel_index(minindex_1d, latvals.shape)
return iy_min,ix_min
ncfile = netCDF4.Dataset('E:\wind_level2_1.nc', 'r')
latvar = ncfile.variables['latitude']
lonvar = ncfile.variables['longitude']
#_________GG turbine_________GAD10 Latitude 51.735516, GAD10 Longitude 1.942656
iy,ix = tunnel_fast(latvar, lonvar, 51.735516, 1.942656)
print('Closest lat lon:', latvar[iy,ix], lonvar[iy,ix])
refLAT=latvar[iy,ix]
refLON = lonvar[iy,ix]
#try to find the data for this location
SARwind = ncfile.variables['sar_wind'][:,:]
ModelWind = ncfile.variables['model_speed'][:,:]
print 'iy,ix' #appears to be the index of the value of Lat,lon
print SARwind[iy,ix]
ncfile.close()
现在我正在尝试遍历包含坐标 coord_list 的文本文件以提取坐标集，找到数据然后移动到列表中的下一组坐标。此代码可独立运行，如下所示：
import csv
from decimal import Decimal
with open('Turbine_locs_no_header.csv','rb') as f:
reader = csv.reader(f)
#coord_list = list(reader)
coord_list = [reader]
end_row = len(coord_list)
lon_ind=1
lat_ind=2
for row in range(0, end_row-1):#end_row - 1 due to the 0 index
turbine_lat = coord_list[row][lat_ind]
turbine_lon = coord_list[row][lon_ind]
turbine_lat = [Decimal(turbine_lat)]
print 'lat',turbine_lat, 'lon',turbine_lon, row
但是，我想将文本文件中的坐标传递给这部分原始代码iy,ix = tunnel_fast(latvar, lonvar, 51.94341, 1.922094888)，用变量iy, ix = tunnel_fast(latvar, lonvar, turbine_lat, turbine_lon)替换数字。我尝试通过创建一个函数 get_coordinates 来组合这两个代码，我得到以下错误
File "C:/Users/mm/test_nc_bycoords_GG_turbines_AGW.py", line 65, in <module>
get_coordinates(coord_list, latvar, lonvar)
File "C:/Users/mm/test_nc_bycoords_GG_turbines_AGW.py", line 51, in get_coordinates
iy, ix = tunnel_fast(latvar, lonvar, turbine_lat, turbine_lon)
File "C:/Users/mm/test_nc_bycoords_GG_turbines_AGW.py", line 27, in tunnel_fast
lat0_rad = lat0 * rad_factor
TypeError: can't multiply sequence by non-int of type 'float'
我认为这是因为 turbine_lat 和 turbine_lon 是列表项所以不能使用，但这似乎与错误无关。我知道这段代码无论如何都需要做更多的工作，但如果有人能帮助我找出我出错的地方，那将非常有帮助。我尝试结合这两个代码如下。
import numpy as np
import netCDF4
from math import pi
from numpy import cos, sin
import csv
# edited from https://github.com/Unidata/unidata-python-workshop/blob/a56daa50d7b343c7debe93968683613642d6b9f7/notebooks/netcdf-by-coordinates.ipynb
def tunnel_fast(latvar,lonvar,lat0,lon0):
'''
Find closest point in a set of (lat,lon) points to specified point
latvar - 2D latitude variable from an open netCDF dataset
lonvar - 2D longitude variable from an open netCDF dataset
lat0,lon0 - query point
Returns iy,ix such that the square of the tunnel distance
between (latval[it,ix],lonval[iy,ix]) and (lat0,lon0)
is minimum.
'''
rad_factor = pi/180.0 # for trignometry, need angles in radians
# Read latitude and longitude from file into numpy arrays
latvals = latvar[:] * rad_factor
lonvals = lonvar[:] * rad_factor
ny,nx = latvals.shape
lat0_rad = lat0 * rad_factor
lon0_rad = lon0 * rad_factor
# Compute numpy arrays for all values, no loops
clat,clon = cos(latvals),cos(lonvals)
slat,slon = sin(latvals),sin(lonvals)
delX = cos(lat0_rad)*cos(lon0_rad) - clat*clon
delY = cos(lat0_rad)*sin(lon0_rad) - clat*slon
delZ = sin(lat0_rad) - slat;
dist_sq = delX**2 + delY**2 + delZ**2
minindex_1d = dist_sq.argmin() # 1D index of minimum element
iy_min,ix_min = np.unravel_index(minindex_1d, latvals.shape)
return iy_min,ix_min
#________________my edits___________________________________________________
def get_coordinates(coord_list, latvar, lonvar):
"this takes coordinates from a .csv and assigns them to variables"
end_row = len(coord_list)
lon_ind=1
lat_ind=2
for row in range(0, end_row-1):#end_row - 1 due to the 0 index
turbine_lat = coord_list[row][lat_ind]
turbine_lon = coord_list[row][lon_ind]
iy, ix = tunnel_fast(latvar, lonvar, turbine_lat, turbine_lon)
print('Closest lat lon:', latvar[iy, ix], lonvar[iy, ix])
#________________________________________________________________________________________________________________________
ncfile = netCDF4.Dataset('NOGAPS_wind_level2_1.nc', 'r')
latvar = ncfile.variables['latitude']
lonvar = ncfile.variables['longitude']
#____added in to pass to get coordinates function
with open('Turbine_locs_no_header.csv','rb') as f:
reader = csv.reader(f)
coord_list = list(reader)
#_________take latitude from coordinateas function
get_coordinates(coord_list, latvar, lonvar)
#iy,ix = tunnel_fast(latvar, lonvar, turbine_lat, turbine_lon)#get these from the 'assign_coordinates_fromlist.py
#print('Closest lat lon:', latvar[iy,ix], lonvar[iy,ix])
SARwind = ncfile.variables['sar_wind'][:,:]
ModelWind = ncfile.variables['model_speed'][:,:]
print 'iy,ix' #appears to be the index of the value of Lat,lon
print SARwind[iy,ix]
ncfile.close()
当我尝试转换时
您可以使用 *args 解压缩参数列表（参见 the docs）。在你的情况下你可以做 tunnel_fast(latvar, lonvar, *coord_list[row])。您需要确保 coord_list[row] 中参数的顺序是正确的，如果 coord_list[row] 包含的值多于这两个值，那么您需要对其进行适当的切片。
感谢 a_guest
的帮助
lat0 和 lon0 被传递为一个简单的问题
当需要 <type 'float'> 时，<type 'str'> 到 tunnel_fast。这似乎来自将 coord_list 作为列表加载。
with open('Turbine_locs_no_header.csv','rb') as f:
reader = csv.reader(f)
coord_list = list(reader)
我使用的解决方法是将 lat0 和 lon0 转换为 tunnel_fast
开头的浮点数
lat0 = float(lat0)
lon0 = float(lon0)
我相信有一种更优雅的方法可以做到这一点，但它确实有效。
我正在使用 workshop 中的一些代码按最接近我指定坐标的坐标从 netCDF 文件中提取数据。当只使用一组坐标时，我可以毫无困难地提取我需要的值，如下所示：
import numpy as np
import netCDF4
from math import pi
from numpy import cos, sin
def tunnel_fast(latvar,lonvar,lat0,lon0):
'''
Find closest point in a set of (lat,lon) points to specified point
latvar - 2D latitude variable from an open netCDF dataset
lonvar - 2D longitude variable from an open netCDF dataset
lat0,lon0 - query point
Returns iy,ix such that the square of the tunnel distance
between (latval[it,ix],lonval[iy,ix]) and (lat0,lon0)
is minimum.
'''
rad_factor = pi/180.0 # for trignometry, need angles in radians
# Read latitude and longitude from file into numpy arrays
latvals = latvar[:] * rad_factor
lonvals = lonvar[:] * rad_factor
ny,nx = latvals.shape
lat0_rad = lat0 * rad_factor
lon0_rad = lon0 * rad_factor
# Compute numpy arrays for all values, no loops
clat,clon = cos(latvals),cos(lonvals)
slat,slon = sin(latvals),sin(lonvals)
delX = cos(lat0_rad)*cos(lon0_rad) - clat*clon
delY = cos(lat0_rad)*sin(lon0_rad) - clat*slon
delZ = sin(lat0_rad) - slat;
dist_sq = delX**2 + delY**2 + delZ**2
minindex_1d = dist_sq.argmin() # 1D index of minimum element
iy_min,ix_min = np.unravel_index(minindex_1d, latvals.shape)
return iy_min,ix_min
ncfile = netCDF4.Dataset('E:\wind_level2_1.nc', 'r')
latvar = ncfile.variables['latitude']
lonvar = ncfile.variables['longitude']
#_________GG turbine_________GAD10 Latitude 51.735516, GAD10 Longitude 1.942656
iy,ix = tunnel_fast(latvar, lonvar, 51.735516, 1.942656)
print('Closest lat lon:', latvar[iy,ix], lonvar[iy,ix])
refLAT=latvar[iy,ix]
refLON = lonvar[iy,ix]
#try to find the data for this location
SARwind = ncfile.variables['sar_wind'][:,:]
ModelWind = ncfile.variables['model_speed'][:,:]
print 'iy,ix' #appears to be the index of the value of Lat,lon
print SARwind[iy,ix]
ncfile.close()
现在我正在尝试遍历包含坐标 coord_list 的文本文件以提取坐标集，找到数据然后移动到列表中的下一组坐标。此代码可独立运行，如下所示：
import csv
from decimal import Decimal
with open('Turbine_locs_no_header.csv','rb') as f:
reader = csv.reader(f)
#coord_list = list(reader)
coord_list = [reader]
end_row = len(coord_list)
lon_ind=1
lat_ind=2
for row in range(0, end_row-1):#end_row - 1 due to the 0 index
turbine_lat = coord_list[row][lat_ind]
turbine_lon = coord_list[row][lon_ind]
turbine_lat = [Decimal(turbine_lat)]
print 'lat',turbine_lat, 'lon',turbine_lon, row
但是，我想将文本文件中的坐标传递给这部分原始代码iy,ix = tunnel_fast(latvar, lonvar, 51.94341, 1.922094888)，用变量iy, ix = tunnel_fast(latvar, lonvar, turbine_lat, turbine_lon)替换数字。我尝试通过创建一个函数 get_coordinates 来组合这两个代码，我得到以下错误
File "C:/Users/mm/test_nc_bycoords_GG_turbines_AGW.py", line 65, in <module>
get_coordinates(coord_list, latvar, lonvar)
File "C:/Users/mm/test_nc_bycoords_GG_turbines_AGW.py", line 51, in get_coordinates
iy, ix = tunnel_fast(latvar, lonvar, turbine_lat, turbine_lon)
File "C:/Users/mm/test_nc_bycoords_GG_turbines_AGW.py", line 27, in tunnel_fast
lat0_rad = lat0 * rad_factor
TypeError: can't multiply sequence by non-int of type 'float'
我认为这是因为 turbine_lat 和 turbine_lon 是列表项所以不能使用，但这似乎与错误无关。我知道这段代码无论如何都需要做更多的工作，但如果有人能帮助我找出我出错的地方，那将非常有帮助。我尝试结合这两个代码如下。
import numpy as np
import netCDF4
from math import pi
from numpy import cos, sin
import csv
# edited from https://github.com/Unidata/unidata-python-workshop/blob/a56daa50d7b343c7debe93968683613642d6b9f7/notebooks/netcdf-by-coordinates.ipynb
def tunnel_fast(latvar,lonvar,lat0,lon0):
'''
Find closest point in a set of (lat,lon) points to specified point
latvar - 2D latitude variable from an open netCDF dataset
lonvar - 2D longitude variable from an open netCDF dataset
lat0,lon0 - query point
Returns iy,ix such that the square of the tunnel distance
between (latval[it,ix],lonval[iy,ix]) and (lat0,lon0)
is minimum.
'''
rad_factor = pi/180.0 # for trignometry, need angles in radians
# Read latitude and longitude from file into numpy arrays
latvals = latvar[:] * rad_factor
lonvals = lonvar[:] * rad_factor
ny,nx = latvals.shape
lat0_rad = lat0 * rad_factor
lon0_rad = lon0 * rad_factor
# Compute numpy arrays for all values, no loops
clat,clon = cos(latvals),cos(lonvals)
slat,slon = sin(latvals),sin(lonvals)
delX = cos(lat0_rad)*cos(lon0_rad) - clat*clon
delY = cos(lat0_rad)*sin(lon0_rad) - clat*slon
delZ = sin(lat0_rad) - slat;
dist_sq = delX**2 + delY**2 + delZ**2
minindex_1d = dist_sq.argmin() # 1D index of minimum element
iy_min,ix_min = np.unravel_index(minindex_1d, latvals.shape)
return iy_min,ix_min
#________________my edits___________________________________________________
def get_coordinates(coord_list, latvar, lonvar):
"this takes coordinates from a .csv and assigns them to variables"
end_row = len(coord_list)
lon_ind=1
lat_ind=2
for row in range(0, end_row-1):#end_row - 1 due to the 0 index
turbine_lat = coord_list[row][lat_ind]
turbine_lon = coord_list[row][lon_ind]
iy, ix = tunnel_fast(latvar, lonvar, turbine_lat, turbine_lon)
print('Closest lat lon:', latvar[iy, ix], lonvar[iy, ix])
#________________________________________________________________________________________________________________________
ncfile = netCDF4.Dataset('NOGAPS_wind_level2_1.nc', 'r')
latvar = ncfile.variables['latitude']
lonvar = ncfile.variables['longitude']
#____added in to pass to get coordinates function
with open('Turbine_locs_no_header.csv','rb') as f:
reader = csv.reader(f)
coord_list = list(reader)
#_________take latitude from coordinateas function
get_coordinates(coord_list, latvar, lonvar)
#iy,ix = tunnel_fast(latvar, lonvar, turbine_lat, turbine_lon)#get these from the 'assign_coordinates_fromlist.py
#print('Closest lat lon:', latvar[iy,ix], lonvar[iy,ix])
SARwind = ncfile.variables['sar_wind'][:,:]
ModelWind = ncfile.variables['model_speed'][:,:]
print 'iy,ix' #appears to be the index of the value of Lat,lon
print SARwind[iy,ix]
ncfile.close()
当我尝试转换时
您可以使用 *args 解压缩参数列表（参见 the docs）。在你的情况下你可以做 tunnel_fast(latvar, lonvar, *coord_list[row])。您需要确保 coord_list[row] 中参数的顺序是正确的，如果 coord_list[row] 包含的值多于这两个值，那么您需要对其进行适当的切片。
感谢 a_guest
的帮助lat0 和 lon0 被传递为一个简单的问题
当需要 <type 'float'> 时，<type 'str'> 到 tunnel_fast。这似乎来自将 coord_list 作为列表加载。
with open('Turbine_locs_no_header.csv','rb') as f:
reader = csv.reader(f)
coord_list = list(reader)
我使用的解决方法是将 lat0 和 lon0 转换为 tunnel_fast
lat0 = float(lat0)
lon0 = float(lon0)
我相信有一种更优雅的方法可以做到这一点，但它确实有效。