将Matlab的datenum格式转换为Python

5 回答

• 15

  使用pandas，可以在尊重小数部分的同时转换整个datenum值数组：

  import numpy as np
  import pandas as pd
  datenums = np.array([737125, 737124.8, 737124.6, 737124.4, 737124.2, 737124])
  timestamps = pd.to_datetime(datenums-719529, unit='D')
  

  值719529是Unix纪元启动（1970-01-01）的datenum值，这是pd.to_datetime()的默认值 origin .

  我使用以下Matlab代码来设置它：

  datenum('1970-01-01')  % gives 719529
  datenums = datenum('06-Mar-2018') - linspace(0,1,6)  % test data
  datestr(datenums)  % human readable format
  

  回复于 2024-04-29T17:00:00+08:00
• 9

  只是 Build 并添加到以前的评论 . 关键是当天计数由 datetime 和类 datetime 中的构造函数 fromordinal 以及相关的子类执行 . 例如，从Python Library Reference for 2.7，一个读取 fromordinal

  返回对应于公历格里高利序数的日期，其中第1年1月1日有序数1.除非1 <= ordinal <= date.max.toordinal（），否则引发ValueError .

  但是，公元0年仍然是一个（跳跃）年份，因此仍有366天需要考虑 . （闰年，就像2016年，恰好是四年前的四十四年 . ）

  这些是我用于类似目的的两个功能：

  import datetime 
  
  def datetime_pytom(d,t):
  '''
  Input
      d   Date as an instance of type datetime.date
      t   Time as an instance of type datetime.time
  Output
      The fractional day count since 0-Jan-0000 (proleptic ISO calendar)
      This is the 'datenum' datatype in matlab
  Notes on day counting
      matlab: day one is 1 Jan 0000 
      python: day one is 1 Jan 0001
      hence an increase of 366 days, for year 0 AD was a leap year
  '''
  dd = d.toordinal() + 366
  tt = datetime.timedelta(hours=t.hour,minutes=t.minute,
                         seconds=t.second)
  tt = datetime.timedelta.total_seconds(tt) / 86400
  return dd + tt
  
  def datetime_mtopy(datenum):
  '''
  Input
      The fractional day count according to datenum datatype in matlab
  Output
      The date and time as a instance of type datetime in python
  Notes on day counting
      matlab: day one is 1 Jan 0000 
      python: day one is 1 Jan 0001
      hence a reduction of 366 days, for year 0 AD was a leap year
  '''
  ii = datetime.datetime.fromordinal(int(datenum) - 366)
  ff = datetime.timedelta(days=datenum%1)
  return ii + ff
  

  希望这有助于并且乐于纠正 .

  回复于 2024-04-29T17:00:00+08:00
• 3

  您链接到解决方案，它有一个小问题 . 就是这个：

  python_datetime = datetime.fromordinal(int(matlab_datenum)) + timedelta(days=matlab_datenum%1) - timedelta(days = 366)
  

  更长的解释可以找到here

  回复于 2024-04-29T17:00:00+08:00
• 2

  为了防止对其他人有用，这里有一个从Matlab mat文件加载时间序列数据的完整示例，使用carlosdc的答案（定义为函数）将Matlab datenums的向量转换为日期时间对象列表，然后绘制为与熊猫的时间序列：

  from scipy.io import loadmat
  import pandas as pd
  import datetime as dt
  import urllib
  
  # In Matlab, I created this sample 20-day time series:
  # t = datenum(2013,8,15,17,11,31) + [0:0.1:20];
  # x = sin(t)
  # y = cos(t)
  # plot(t,x)
  # datetick
  # save sine.mat
  
  urllib.urlretrieve('http://geoport.whoi.edu/data/sine.mat','sine.mat');
  
  # If you don't use squeeze_me = True, then Pandas doesn't like 
  # the arrays in the dictionary, because they look like an arrays
  # of 1-element arrays.  squeeze_me=True fixes that.
  
  mat_dict = loadmat('sine.mat',squeeze_me=True)
  
  # make a new dictionary with just dependent variables we want
  # (we handle the time variable separately, below)
  my_dict = { k: mat_dict[k] for k in ['x','y']}
  
  def matlab2datetime(matlab_datenum):
      day = dt.datetime.fromordinal(int(matlab_datenum))
      dayfrac = dt.timedelta(days=matlab_datenum%1) - dt.timedelta(days = 366)
      return day + dayfrac
  
  # convert Matlab variable "t" into list of python datetime objects
  my_dict['date_time'] = [matlab2datetime(tval) for tval in mat_dict['t']]
  
  # print df
  <class 'pandas.core.frame.DataFrame'>
  DatetimeIndex: 201 entries, 2013-08-15 17:11:30.999997 to 2013-09-04 17:11:30.999997
  Data columns (total 2 columns):
  x    201  non-null values
  y    201  non-null values
  dtypes: float64(2)
  
  # plot with Pandas
  df = pd.DataFrame(my_dict)
  df = df.set_index('date_time')
  df.plot()
  

  

  回复于 2024-04-29T17:00:00+08:00
• 2

  这是使用 numpy.datetime64 而不是 datetime 转换它们的方法 .

  origin = np.datetime64('0000-01-01', 'D') - np.timedelta64(1, 'D')
  date = serdate * np.timedelta64(1, 'D') + origin
  

  这适用于 serdate 单个整数或整数数组 .

  回复于 2024-04-29T17:00:00+08:00