Thanks a lot Pauli!
Best,
Chao
2011/9/29 Pauli Virtanen
> 29.09.2011 17:38, Benjamin Root kirjoitti:
> [clip]
> > but I cannot use "d=np.concatenate(d,np.array(np.nan))" to finish this
> job.
> [clip]
>
> Do
>
>d = np.concatenate([d, np.array([np.nan])])
>
> or
>
>d = np.hstack([d, np.
Thanks Ben! it works fine! help me a lot.
Chao
2011/9/29 Pauli Virtanen
> 29.09.2011 17:38, Benjamin Root kirjoitti:
> [clip]
> > but I cannot use "d=np.concatenate(d,np.array(np.nan))" to finish this
> job.
> [clip]
>
> Do
>
>d = np.concatenate([d, np.array([np.nan])])
>
> or
>
>d = np
29.09.2011 17:38, Benjamin Root kirjoitti:
[clip]
> but I cannot use "d=np.concatenate(d,np.array(np.nan))" to finish this job.
[clip]
Do
d = np.concatenate([d, np.array([np.nan])])
or
d = np.hstack([d, np.nan])
or
d = np.r_[d, np.nan]
---
On Thu, Sep 29, 2011 at 9:33 AM, Chao YUE wrote:
> Dear all, I have a variable d which has several years plus 11 month data
> (len(d)%12=11). so I want to append a NaN to the data so that it constitutes
> complete
> several years of data.
>
> but I cannot use "d=np.concatenate(d,np.array(np.nan))
