您可以通過(guò)迭代字典中的每個(gè)項(xiàng)目來(lái)做到這一點(diǎn)，但是在大型數(shù)據(jù)集中它可能效率低下。def get_key(data, query):    for key, value in data.items():        if query in value[0][1]:            return key    return 'Not Found'get_key(dictonary, word)然后，即使您的查找未能找到匹配項(xiàng)，您也可以調(diào)用您的函數(shù)并返回結(jié)果。# Note i changed the name of the dictionary to dicton, as dict shouldn't be used as a variable nameprint(get_key(dicton, 'three'))print(get_key(dicton, 'seven'))print(get_key(dicton, 'four'))#a#Not Found#b我有一些 0.2 度和 1 度分辨率的虛擬數(shù)據(jù)。我想對(duì) foo 進(jìn)行二次采樣，使其與 foo1 的比例相同。有什么簡(jiǎn)單的方法可以以某種方式平均和重新排列我的經(jīng)緯度坐標(biāo)嗎？import pandas as pdimport xarray as xrimport matplotlib.pyplot as plt#Set at 0.2 degree grids ishfreq=20lats=240lons=1020time=pd.date_range('2000-01',periods=freq,freq='Y')data=np.random.rand(freq,lats,lons)lat=np.linspace(-19.5,19.5,lats)lon=np.linspace(120,290,lons)foo = xr.DataArray(data, coords=[time, lat,lon], dims=['time', 'lat','lon'])foo.sel(time='2005',method='nearest').plot()plt.show()#Set at 1 degree gridsfreq1=20lats1=40 #Factor of 6 differencelons1=170time1=pd.date_range('2000-01',periods=freq1,freq='Y')data1=np.random.rand(freq1,lats1,lons1)lat1=np.linspace(-19.5,19.5,lats1)lon1=np.linspace(120,290,lons1)foo1 = xr.DataArray(data1, coords=[time1, lat1,lon1], dims=['time', 'lat','lon'])foo1.sel(time='2005',method='nearest').plot()plt.show()