Matlab struct数组转换为python [英] Matlab struct array to python

问题描述

我已经仔细研究了一下，试图找出一种在python中创建类似matlab的struct数组的方法.我在网上查看了一些问题，答案似乎没有帮助，或者我可能只是误解了它们，因为它们与我有关.因此，继续前进.我正在尝试形成一个等效于以下matlab代码的python.

I have thoroughly looked around to try figuring out a way to create a matlab like struct array in python. There are some questions online that I looked at and either the answers don't seem to help or I might simply be misinterpreting them as they pertain to me. So, moving on. I am trying to form a python equivalent to the following matlab code.

channel                 = [];
channel.PRN             = 0; 
channel.acquiredFreq    = 0; 
channel.codePhase       = 0; 
channel.status          = '-';  
channel = repmat(channel, 1, settings.numberOfChannels); 

其中repmat基本上会创建一个名为channel的结构数组，其中的单元数等于settings.numberOfChannels，并且每个单元都有PRN，acquiredFreq等.

Where repmat would basically create a struct array called channel with a number of cells equal to settings.numberOfChannels and each of those would have PRN,acquiredFreq, etc.

稍后，我通过执行循环来更改这些值，从而访问此结构:

Later on, I access this struct by performing a loop that alters these values as such:

for ii = 1:settings.numberOfChannels
        channel(ii).PRN          = PRNindexes(ii);
        channel(ii).acquiredFreq = acqResults.carrFreq(PRNindexes(ii));
        channel(ii).codePhase    = acqResults.codePhase(PRNindexes(ii));

我尝试了几种方法，但是在使用numpy的tile情况下(我可能一直使用不正确)，它会吐出一些废话，或者当我尝试进行如下循环时:

I have tried several approaches but it either spits out nonsense in the case of tile using numpy(which i might have just been using incorrectly) or when I attempt to make a loop such as:

class test:
    for iii in range(1,settings.numberOfChannels):
        iii.PRN=0
        iii.acquiredFreq=0
        iii.codePhase=0
        iii.status="-"

很有可能我认为这是语法错误或对python的误解，因为这是我第一次使用它.如果这是问这个问题或类似问题的不正确地点，我深表歉意.

More than likely I suppose it's a syntax error or my misunderstanding of python since this is my first time utilizing it. If this is the incorrect place to ask this or something of that nature, I apologize.

谢谢

推荐答案

更新:您可能想调查熊猫.它的 与NumPy相比，Series和DataFrames更易于使用且功能更全 结构化数组.

Update: You may want to investigate Pandas. Its Series and DataFrames are easier to work with and more full-featured than NumPy structured arrays.

您可以使用NumPy 结构化数组:

You could use a NumPy structured array:

import numpy as np
channel = np.zeros(1, dtype = [('PRN',int),
                               ('acquiredFreq',int),
                               ('codePhase',int),
                               ('status','|S1')])

print(channel)
# [(0, 0, 0, '')]

按整数索引访问特定行:

Indexing by integer accesses a particular row:

print(channel[0])
# (0, 0, 0, '')

按列名索引将列作为数组返回:

Indexing by column name returns the column as an array:

print(channel['PRN'])
# [0]

或者您可以遍历每一行并分配给每个字段(列)， 但这在NumPy中相对较慢.

Or you can loop through each row and assign to each field (column), but this is relatively slow in NumPy.

for row in channel:
    row['PRN'] = 1
    row['acquiredFreq'] = 1
    row['codePhase'] = 1
    row['status'] = '+'

print(channel)    
# [(1, 1, 1, '+')]

为了完整起见，我还要提到您可以按行然后按列进行分配:

Just for completeness, I'll also mention you can assign by row then column:

channel[0]['status'] = '-'
print(channel)
# [(1, 1, 1, '-')]

或按列分配，然后按行分配:

or assign by column then row:

channel['PRN'][0] = 10
print(channel)
# [(10, 1, 1, '-')]

我展示了上面的内容，因为它与您发布的Matlab代码最相似.但是，让我再次强调，分配给NumPy数组中的单个单元很慢. NumPy执行上述操作的方法是执行整个数组分配:

I showed the above because it is most similar to the Matlab code you posted. However, let me stress again that assigning to individual cells in a NumPy array is slow. The NumPy-way to do the above is to do whole-array assignments instead:

channel['PRN'] = PRNindexes

其中PRNindexes是一个序列(例如列表，元组或NumPy数组).

where PRNindexes is a sequence (e.g. a list, tuple, or NumPy array).

您还可以使用花式索引(又称高级索引")以选择行:

You can also use fancy indexing (aka "advanced indexing") to select rows:

index = (channel.status == '+')  # Select all rows with status '+'
channel['PRN'][index] = 10       # Set PRN to 10 for all those rows

请记住，精美索引会返回一个新数组，而不是原始数组的视图. (相反，基本切片"(例如channel[0]或channel[1:10]返回一个视图.)因此，如果要进行分配来更改原始数组的赋值，请先按列选择，然后按花式索引(index)

Just keep in mind that fancy indexing returns a new array, not a view of the original array. (In contrast, "basic slicing" (e.g. channel[0] or channel[1:10] returns a view.) So if you want to make assignments that alter the original array, select by column first, then fancy index (index)

channel['PRN'][index] = ...

而不是

channel[index]['PRN'] = ...

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