如何解决我想知道,如何绘制异常海面温度的 T 检验结果? (SST) 数据文件 = sst.mean.anom.nc
我想知道,如何绘制异常海面温度 (SST) 数据文件 = sst.mean.anom.nc 的 T 检验结果。 我写了两个代码,但我不知道这两个代码哪个效果更好。
代码1
meanSumaX = sumaX / len(X)
meanSumaHx = sumaHx / len(Hx)
difMeanSumaHxX = meanSumaHx - meanSumaX
ddof_MeanSumaHx = meanSumaHx.var(ddof=1)
ddof_MeanSumaX = meanSumaX.var(ddof=1)
s = np.sqrt((ddof_MeanSumaX + ddof_MeanSumaHx)/2)
t = difMeanSumaHxX / (s * np.sqrt(2/len(sst.loc['1948-03-01':'1998-03-01':12])))
# Using geophysical units. `robust` disregards outliers for colour map creation.
fig = plt.figure(5,figsize=(15.,12.))
ax = plt.axes(projection=ccrs.PlateCarree(central_longitude=0.0))
ax.coastlines()
ax.add_feature(cf.LAND) # utilizar con internet
t.plot()
plt.title("Mean SST anomali (C) 1948-03 - 1998-03 T - test ")
ax.gridlines(draw_labels=True)
plt.show()
代码 2
meanSumaX = sumaX / len(X)
meanSumaHx = sumaHx / len(Hx)
difMeanSumaHxX = meanSumaHx - meanSumaX
std1,std2 = std(meanSumaHx,ddof=1),std(meanSumaX,ddof=1)
se1,se2 = std1 / np.sqrt(len(meanSumaHx)),std2 / np.sqrt(len(meanSumaX))
sed = np.sqrt(se1 ** 2 + se2 ** 2)
t = difMeanSumaHxX / sed
# Using geophysical units. `robust` disregards outliers for colour map creation.
fig = plt.figure(5,12.))
ax = plt.axes(projection=ccrs.PlateCarree(central_longitude=0.0))
ax.coastlines()
ax.add_feature(cf.LAND) # utilizar con internet
t.plot()
plt.title("Mean SST anomali (C) 1948-03 - 1998-03 T - test ")
ax.gridlines(draw_labels=True)
plt.show()
版权声明:本文内容由互联网用户自发贡献,该文观点与技术仅代表作者本人。本站仅提供信息存储空间服务,不拥有所有权,不承担相关法律责任。如发现本站有涉嫌侵权/违法违规的内容, 请发送邮件至 dio@foxmail.com 举报,一经查实,本站将立刻删除。