我正在使用按位左移运算符创建一个numpy数组。
例如,我创建数组p,其中数组的形状与矩阵a的形状相同,即(23,):
>>> import numpy
>>> a = numpy.array([0,0,0,0,0,0,1,0,1,0,1,1,1,0,0,0,0,0,1,0,0,1,1])
>>> p = 1 << arange(a.shape[-1] - 1) #left shift
结果如预期:
>>> p
array([ 1, 2, 4, 8, 16, 32, 64,
128, 256, 512, 1024, 2048, 4096, 8192,
16384, 32768, 65536, 131072, 262144, 524288, 1048576,
2097152])
但是,如果我们增加数组的大小,让我们说(70,):
>>> a = numpy.array([0,0,0,0,0,0,1,0,1,0,1,1,1,0,0,0,0,0,1,0,0,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0])
>>> p = 1 << arange(a.shape[-1] - 1, -1, -1) #left shift
>>> p
array([ 1, 2, 4,
8, 16, 32,
64, 128, 256,
512, 1024, 2048,
4096, 8192, 16384,
32768, 65536, 131072,
262144, 524288, 1048576,
2097152, 4194304, 8388608,
16777216, 33554432, 67108864,
134217728, 268435456, 536870912,
1073741824, 2147483648, 4294967296,
8589934592, 17179869184, 34359738368,
68719476736, 137438953472, 274877906944,
549755813888, 1099511627776, 2199023255552,
4398046511104, 8796093022208, 17592186044416,
35184372088832, 70368744177664, 140737488355328,
281474976710656, 562949953421312, 1125899906842624,
2251799813685248, 4503599627370496, 9007199254740992,
18014398509481984, 36028797018963968, 72057594037927936,
144115188075855872, 288230376151711744, 576460752303423488,
1152921504606846976, 2305843009213693952, 4611686018427387904,
-9223372036854775808, 0, 0,
0, 0, 16])
在顶部你可以看到,当它从1,2,4,8增加时,.....变为负数,然后是0,最后是16。
如果我单独进行,则不是这种情况:
>>> 1<<70
1180591620717411303424
那么,我该怎么做才能使我的数组元素的值与1<<x
相对应,其中x是一个高数字(大于70)?
Python int和numpy int不一样...... Python支持任意长度,而numpy由类型修复:
numpy.array([1]) << 70
>>> array([64], dtype=int32)
一种解决方案是使用对象dtype:
numpy.array([1], dtype=numpy.object) << 70
>>> array([1180591620717411303424], dtype=object)
以下内容将按预期工作:
a = numpy.array([1], dtype=numpy.object) << numpy.arange(70)
查看最后一个元素的类型,我们看到它是一个Python int:
type(a[-1])
>>> <class 'int'>