NCL中EOF问题

风之牧语

在使用NCL中自带的函数eofunc分解空间模态的时候，发现使用原值和距平分解出来的空间模态和时间序列是一样的。请问这是什么问题或者说自带函数已经默认将原值求解为了距平场？思考了好些天了，至今未解，跪求大神们解惑啊！

随缘

Description

This function computes Empirical Orthogonal Functions (EOFs) via a covariance matrix or, optionally, via a correlation matrix. This is also known as Principal Component Analysis or Eigen Analysis. The EOFs are calculated using LAPACK's "dspevx" routine. Missing values are ignored when computing the covariance or correlation matrix. The returned values are normalized such that the sum of squares for each EOF pattern equals one. To denormalize the returned EOFs multiply by the square root of the associated eigenvalue (aka, the singular value).

If data does NOT have time as the rightmost dimension, then use eofunc_n to avoid having to reorder the data.

Most commonly, the input data consists of anomalies.

This function differs from the deprecated eofcov and eofcor functions in that it may transpose the input data array prior to computing the EOFs. If data is transposed, a linear transformation is applied to the EOFs of the transposed array prior to returning. The reason for using this approach is computational efficiency.

Comments on weighting observations

Generally, when performing and EOF analysis on observations over the globe or a portion of the globe, the values are weighted prior to calculating. This is usually required to account for the convergence of the meridions (area weighting) which lessens the impact of high-latitude grid points that represent a small area of the globe. Most frequently, the square root of the cosine of the latitude is used to compute the area weight. The square root is used to create a covariance matrix that reflects the area of each matrix element. If weighted in this manner, the resulting covariance values will include quantities calculated via:

[x*sqrt(cos(lat(x)))]*[y*sqrt(cos(lat(y)))] = x*y*sqrt(cos(lat(x)))*sqrt(cos(lat(y)))
Note that the covariance of a grid point with itself yields standard cosine weighting:
o

[x*sqrt(cos(lat(x)))]*[x*sqrt(cos(lat(x)))] = x^2 * cos(lat(x)).
Note on standard EOF analysis
Conventional EOF analysis yields patterns and time series which are both orthogonal. The derived patterns are a function of the domain. However, the EOF procedure is strictly mathematical (not statistical) and is not based upon physics. The results may produce patterns that are similar to physical modes within the the system. However, physical meaning is dependent on your interpretation of the mathematical result.

Note on signs of EOF analysis (conributed by Andrew Dawson, UEA)

EOFs are eigenvectors of the covariance matrix formed from the input data. Since an eigenvector can be multiplied by any scalar and still remain an eigenvector, the sign is arbitrary. In a mathematical sense the sign of an eigenvector is rather unimportant. This is why the EOF analysis may yield different signed EOFs for slightly different inputs. Sign only becomes an issue when you wish to interpret the physical meaning (if any) of an eigenvector.

You should approach the interpretation of EOFs by looking at both the EOF pattern and the associated time series together. For example, consider an EOF of sea surface temperature. If your EOF has a positive centre and the associated time series is increasing, then you will interpret this centre as a warming signal. If your EOF had come out the other sign (ie. a negative centre), then the associated time series would also be the opposite sign and you would still interpret the centre as a warming signal.

In essence, the sign flip does not change the physical interpretation of the result. Hence, it is up to you to choose which sign to associate with your EOF patterns for visualisation (remembering that any sign change to an EOF must be applied to the associated time series also). Usually you would simply adjust the sign so that all your EOF patterns with the same physical interpretation also look the same.

If desired, EOF spatial patterns may be tested for orthogonality by using the dot product:

  d01 = sum(eof(0,:,:)*eof(1,:,:))
  d12 = sum(eof(1,:,:)*eof(2,:,:))
  d02 = sum(eof(0,:,:)*eof(2,:,:))
  print("d01="+d01+"  d12="+d12+"  d02="+d02)  ; may be +/- 1e-8

Use eofunc_Wrap if retention of metadata is desired.
References:


Quadrelli, Roberta, Christopher S. Bretherton, John M. Wallace, 2005:
On Sampling Errors in Empirical Orthogonal Functions.
J. Climate, 18, 3704-3710
     
North, G. R., T. L. Bell, R. F. Cahalan, and F. J. Moeng, Sampling
errors in the estimation of empirical orthogonal functions, Mon.
Wea. Rev., 110, 699-706, 1982.
   
Dawson, A.: EOF Analysis
Acknowledgement: The code used is a modified version of David Pierce's fortran code.

随缘

Most commonly, the input data consists of anomalies.

sf2014xiafei

厉害，很值得学习

风之牧语

随缘 发表于 2017-4-12 10:13
Most commonly, the input data consists of anomalies.

原谅我反应慢，还是没明白，通常是用距平做EOF。但是我想知道距平是NCL自带EOF函数会自动处理了还是自己先处理好？

藕茭白空心菜

风之牧语 发表于 2017-4-13 16:38
原谅我反应慢，还是没明白，通常是用距平做EOF。但是我想知道距平是NCL自带EOF函数会自动处理了还是自己 ...

请问楼主知道了吗~

藕茭白空心菜

风之牧语 发表于 2017-4-13 16:38
原谅我反应慢，还是没明白，通常是用距平做EOF。但是我想知道距平是NCL自带EOF函数会自动处理了还是自己 ...

我觉得是自己算好距平再做EOF

紫藤萝

应该是NCL自带做距平了吧，我算出来的也是一样的

紫藤萝

求懂的人指导一下

liweier123

应该是系统直接处理，只有距平值才能算协方差