我想从 MDF 文件转换为 Dataframe

Question

我使用像 asammdf 或 mdfreader 这样的 MDF 文件。但是我的数据文件太大了。所以，我想读取特定数据并制作一个数据框。mdf 数据读取器使用 asammdf from asammdf import MDFcase in mdfreader，有很多错误，因为我的 mdf 文件是同名数据，重采样时有些麻烦（错误数据出来）filename = test_t16.dat" ; 我的 MDF 数据文件yop = MDF(文件名); 使用 asammdf 的 mdf 阅读器whl_rr = yop.get('WHL_SPD_RR') ;我从 mdf 文件中选取一些数据(yop)whl_rr =invalidation_bits = {NoneType} 无master_metadata = {tuple} ('TimeChannel', 1)name = {str} 'WHL_SPD_RR'原始 = {bool} 假样本 = {ndarray} [0. ... 0.]来源 = {NoneType} 无stream_sync = {bool} 假时间戳 = {ndarray} [ 240.4053 ... 2050.81525]unit = {str} 'km/h' > 我想要这样的数据框from asammdf import MDFimport numpy as npimport pandas as pdimport matplotlib.pyplot as pltfilename = r"C:\Users\wonyo\PycharmProjects\test\test_t16.dat"yop = MDF(filename)signallist = [ "WHL_SPD_RR","WHL_SPD_FR", "WHL_SPD_RL","WHL_SPD_FL"]df=[]def group_len(yop, start, stop):                 for i in range(start, stop):                   if yop.get_group(i).empty != True:            if i == start:                max_len = len(yop.get_group(i).TimeChannel)                min_time = min(yop.get_group(i).TimeChannel)            else:                max_len = min(max_len, len(yop.get_group(i).TimeChannel))                min_time = max(min_time, min(yop.get_group(i).TimeChannel))    return max_len-1000, max(242, min_time + 2) grlen_time = group_len(yop, 68, 140)max_len = grlen_time[0]min_time = grlen_time[1]time = np.linspace(0, (max_len - 1) * 0.01, max_len)for i in range(0,5):    signal = yop.get(signallist[i])    signal.timestamps = signal.timestamps - min_time    signal = signal.interp(time)    data_sg = signal.samples    name_sg = signal.name    inex_sg = signal.timestamps    mydata = pd.DataFrame( data =data_sg , index=index_sg, columns=name_sg)    print(df)whl_spd_fl = signal()

Answer 1

df= mdf.filter(signallist).export(fmt='pandas')

Answer 2

yop = MDF(file_name, memory='minimum')

to_keep = ['Channl1', 'Channel2', ('CR_Bms_Soc_Pc', 102) ] # and so on

df = yop.filter(to_keep).cut(start=240, stop=2050).export('pandas', raster=0.01)