将字段添加到结构化numpy数组的最简洁方法是什么?它可以破坏性地完成,还是需要创建一个新的数组并复制现有的字段?每个字段的内容是否连续存储在内存中,以便可以有效地完成这种复制?
答案 0 :(得分:19)
如果您正在使用numpy 1.3,那么还有numpy.lib.recfunctions.append_fields()。
对于许多安装,您需要import numpy.lib.recfunctions
才能访问此内容。 import numpy
不允许用户查看numpy.lib.recfunctions
答案 1 :(得分:6)
import numpy
def add_field(a, descr):
"""Return a new array that is like "a", but has additional fields.
Arguments:
a -- a structured numpy array
descr -- a numpy type description of the new fields
The contents of "a" are copied over to the appropriate fields in
the new array, whereas the new fields are uninitialized. The
arguments are not modified.
>>> sa = numpy.array([(1, 'Foo'), (2, 'Bar')], \
dtype=[('id', int), ('name', 'S3')])
>>> sa.dtype.descr == numpy.dtype([('id', int), ('name', 'S3')])
True
>>> sb = add_field(sa, [('score', float)])
>>> sb.dtype.descr == numpy.dtype([('id', int), ('name', 'S3'), \
('score', float)])
True
>>> numpy.all(sa['id'] == sb['id'])
True
>>> numpy.all(sa['name'] == sb['name'])
True
"""
if a.dtype.fields is None:
raise ValueError, "`A' must be a structured numpy array"
b = numpy.empty(a.shape, dtype=a.dtype.descr + descr)
for name in a.dtype.names:
b[name] = a[name]
return b