Closed HuixuanGan closed 5 months ago
from asammdf import MDF
m = MDF(r'c:\ASAP2_Demo_V171.mf4')
for i, df in enumerate(m.iter_groups()):
print(df)
print(df.index)
ASAM_[15].M.MATRIX_DIM_16.UBYTE.IDENTICAL ASAM.M.SCALAR.UBYTE.FORM_X_PLUS_4
timestamps
0.000000 244 29.0
0.010000 247 32.0
0.020000 250 34.0
0.030000 253 37.0
0.040000 0 39.0
... ... ...
12.350015 229 166.0
12.360015 230 164.0
12.370015 230 163.0
12.380015 230 161.0
12.390015 230 159.0
[1240 rows x 2 columns]
Index([ 0.0, 0.009999999999990905, 0.01999999999998181,
0.03000000000002956, 0.040000000000020464, 0.05000000000001137,
0.060000000000002274, 0.06999999999999318, 0.07999999999998408,
0.09000000000003183,
...
12.300015452548678, 12.310015442978454, 12.320015433372419,
12.330015423730401, 12.340015414052516, 12.350015404338649,
12.360015394588913, 12.370015384803253, 12.380015374981667,
12.390015365124214],
dtype='float64', name='timestamps', length=1240)
the timestamps are found in the dataframe index
Thanks!
Could anyone tell me how to extrace all the timestamps as epoch form inside of the dataframe?
My current reading of timestamps from channel groups using: mdf.channels_db only include results like this: {'Timestamp': ((0, 0), (1, 0), (2, 0), (3, 0), (4, 0), (5, 0), (6, 0), (7, 0), (8, 0), (10, 0), (12, 0), (14, 0), (16, 0), (17, 0), (18, 0), (19, 0), (20, 0), (21, 0), (22, 0), (23, 0), (24, 0), (25, 0), (26, 0), (27, 0), (28, 0), (29, 0), (31, 0), (33, 0), (35, 0), (37, 0), (38, 0), (39, 0)} that may determine where have the timestamp, not the timestamp itself.
Also the iter_groups() doesn't have timestamp as a column.
Thank you!