将 16 位整数转换为 32 位浮点数
我正在尝试将用另一种语言（一种由 Wavemetrics 开发的名为 Igor Pro 的晦涩难懂的语言，你们可能听说过它）编写的代码的一部分移植到 Python 中。
在此代码中，数据类型从 16 位整数（从 16 位大端二进制文件读取）转换为单精度（32 位）浮点。在该程序中，转换如下：
有符号 16 位整数：
print tmp
tmp[0]={-24160,18597,-24160,18597,-24160}
转换为 32 位浮点数：
Redimension/S/E=1 tmp
print tmp
tmp[0]={339213,339213,5.79801e-41,0,0}
/S 标志/选项表示 tmp 应该是 float32 而不是 int16。但是，我相信重要的标志/选项是 /E=1，据说“强制重塑而不转换或移动数据”。
在Python中，转换如下：
>>> tmp[:5]
array([-24160, 18597, -24160, 18597, -24160], dtype=int16)
>>> tmp.astype('float32')
array([-24160., 18597., -24160., ..., 18597., -24160., 18597.], dtype=float32)
这是我所期望的，但我需要找到一个模拟上面原始代码中的/E=1选项的函数/操作。有没有一种明显的方法可以将 -24160 和 18597 都转换为 339213 ？这与 byteswap 或 newbyteorder 或其他什么有关系吗？
I am trying to port a portion of a code written in a different language (an obscure one called Igor Pro by Wavemetrics for those of you have heard of it) to Python.
In this code, there is a conversion of a data type from a 16-bit integer (read in from a 16-bit, big endian binary file) to single-precision (32-bit) floating-point. In this program, the conversion is as follows:
Signed 16-bit integer:
print tmp
tmp[0]={-24160,18597,-24160,18597,-24160}
converted to 32-bit floating-point:
Redimension/S/E=1 tmp
print tmp
tmp[0]={339213,339213,5.79801e-41,0,0}
The /S flag/option indicates that the data type of tmp should be float32 instead of int16. However, I believe the important flag/option is /E=1, which is said to "Force reshape without converting or moving data."
In Python, the conversion is as follows:
>>> tmp[:5]
array([-24160, 18597, -24160, 18597, -24160], dtype=int16)
>>> tmp.astype('float32')
array([-24160., 18597., -24160., ..., 18597., -24160., 18597.], dtype=float32)
Which is what I expect, but I need to find a function/operation that emulates the /E=1 option in the original code above. Is there an obvious way in which -24160 and 18597 would both be converted to 339213? Does this have anything to do with byteswap or newbyteorder or something else?
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结果：
我必须在值列表中添加一个零，因为值的数量为奇数。它无法将这些解释为 32 位浮点数，因为有 5 个 16 位值。
Result:
I had to add a zero to the list of value because there are an odd number of values. It cannot interpret those as 32 bit floats since there 5 16 bit values.
使用 view 而不是 astype:
.astype创建一个副本数组的转换为新的dtype.view返回数组的视图（具有相同的基础数据），根据新的数据类型解释数据。
Use view instead of astype:
.astypecreates a copy of the array cast to the new dtype.viewreturns a view of the array (with the same underlying data),with the data interpreted according to the new dtype.
不，但也没有任何明显的方法可以将 -24160 转换为 339213 和 5.79801e-41 和 0。
看起来更像是转换需要两个 输入数字来创建一个 输出（可能通过将原始 2×16 位连接到 32 位并将结果称为浮点数）。在这种情况下，
-24160,18597对始终变为 339213，而 5.79801e-41 可能是由-24160,0产生的，其中 0 是因为我们用完输入而发明的。由于 5.79801e-41 看起来可能是单精度非正规，这意味着两个 16 位块可能以小端顺序连接。是否需要对每个 16 位输入进行字节交换还有待观察，但您可以自己检查。
No, but neither is there any obvious way in which -24160 would convert to 339213 and 5.79801e-41 and 0.
It looks more like the conversion takes two input numbers to create one output (probably by concatenating the raw 2×16 bits to 32 bits and calling the result a float). In that case the pair
-24160,18597consistently becomes 339213, and 5.79801e-41 probably results from-24160,0where the 0 is invented because we run out of inputs. Since 5.79801e-41 looks like it might be a single-precision denormal, this implies that the two 16-bit blocks are probably concatenated in little-endian order.It remains to see whether you need to byte-swap each of the 16-bit inputs, but you can check that for yourself.