python - 用列表理解替换 nan

Question

yu = np.array([np.nan, np.nan, np.nan, np.nan, np.nan,])

rte = np.array([1,2,3,4,5])

yu[0] = rte[0]
yu # array([  1.,  nan,  nan,  nan,  nan])

还..

[yu[i] = rte[i] for i in range(len(rte))]

语法错误:语法无效

具体来说，我试图用其他相同长度的数组填充数组中的 nan:

[pred[first_c_rowNA, 0::][0::, wNA[0]][i] = output[i] for i in np.arange(len(output))]

语法错误:语法无效

pred[first_c_rowNA, 0::][0::, wNA[0]] # array([ nan,  nan,  nan, ...,  nan,  nan,  nan])

和

output # array([ 0.,  0.,  0., ...,  0.,  0.,  0.]) # not all are zeros

谢谢

Answer 1

为什么不简单地像这样:

>>> import numpy as np
>>> rte = np.array([1,2,3,4,5])
>>> yu = np.array([np.nan, np.nan, np.nan, np.nan, np.nan,])
>>> yu[:] = rte

>>> yu
array([ 1.,  2.,  3.,  4.,  5.])

或者如果你需要纳米掩码:

yu[np.isnan(yu)] = values

例如:

>>> yu
array([ 0.20087116,         nan,  0.71742786,  0.05037165,  0.25646742,
               nan,  0.27702335,         nan,         nan,  0.62272575])

>>> yu[np.isnan(yu)] = np.random.rand(4)

>>> yu
array([ 0.20087116,  0.6701011 ,  0.71742786,  0.05037165,  0.25646742,
        0.63462273,  0.27702335,  0.01248758,  0.61178318,  0.62272575])