Error when trying to use shuffle=3

[eawer]

Greetings! Decord version is '0.4.0'

Here is my code:

vl = VideoLoader(videos, ctx=[cpu(0)], shape=(10, 1280, 720, 3), interval=1, skip=5, shuffle=3)

And here is my error:

DECORDErrorTraceback (most recent call last)
<timed exec> in <module>

/opt/conda/lib/python3.6/site-packages/decord/video_loader.py in __init__(self, uris, ctx, shape, interval, skip, shuffle, prefetch)
     53         assert len(shape) == 4, "expected shape: [bs, height, width, 3], given {}".format(shape)
     54         self._handle = _CAPI_VideoLoaderGetVideoLoader(
---> 55             uri, device_types, device_ids, shape[0], shape[1], shape[2], shape[3], interval, skip, shuffle, prefetch)
     56         assert self._handle is not None
     57         self._len = _CAPI_VideoLoaderLength(self._handle)

/opt/conda/lib/python3.6/site-packages/decord/_ffi/_ctypes/function.py in __call__(self, *args)
    173         check_call(_LIB.DECORDFuncCall(
    174             self.handle, values, tcodes, ctypes.c_int(num_args),
--> 175             ctypes.byref(ret_val), ctypes.byref(ret_tcode)))
    176         _ = temp_args
    177         _ = args

/opt/conda/lib/python3.6/site-packages/decord/_ffi/base.py in check_call(ret)
     61     """
     62     if ret != 0:
---> 63         raise DECORDError(py_str(_LIB.DECORDGetLastError()))
     64 
     65 

DECORDError: [08:51:44] /io/decord/src/video/video_loader.cc:70: Invalid shuffle mode: 3 Available: 
	{No shuffle: 0}
	{Random File Order: 1}
	{Random access: 2}

Stack trace returned 10 entries:
[bt] (0) /opt/conda/lib/python3.6/site-packages/decord/libdecord.so(dmlc::StackTrace(unsigned long)+0x50) [0x7f3ea0501830]
[bt] (1) /opt/conda/lib/python3.6/site-packages/decord/libdecord.so(dmlc::LogMessageFatal::~LogMessageFatal()+0x1d) [0x7f3ea050291d]
[bt] (2) /opt/conda/lib/python3.6/site-packages/decord/libdecord.so(decord::VideoLoader::VideoLoader(std::vector<std::string, std::allocator<std::string> >, std::vector<DLContext, std::allocator<DLContext> >, std::vector<int, std::allocator<int> >, int, int, int, int)+0xe54) [0x7f3ea054c4c4]
[bt] (3) /opt/conda/lib/python3.6/site-packages/decord/libdecord.so(+0x6d5be) [0x7f3ea05455be]
[bt] (4) /opt/conda/lib/python3.6/site-packages/decord/libdecord.so(DECORDFuncCall+0x52) [0x7f3ea04fe412]
[bt] (5) /opt/conda/lib/python3.6/lib-dynload/../../libffi.so.6(ffi_call_unix64+0x4c) [0x7f3ebcce7ec0]
[bt] (6) /opt/conda/lib/python3.6/lib-dynload/../../libffi.so.6(ffi_call+0x22d) [0x7f3ebcce787d]
[bt] (7) /opt/conda/lib/python3.6/lib-dynload/_ctypes.cpython-36m-x86_64-linux-gnu.so(_ctypes_callproc+0x2ce) [0x7f3ebcefdede]
[bt] (8) /opt/conda/lib/python3.6/lib-dynload/_ctypes.cpython-36m-x86_64-linux-gnu.so(+0x13915) [0x7f3ebcefe915]
[bt] (9) /opt/conda/bin/python(_PyObject_FastCallDict+0x8b) [0x561e814a1e3b]

How can I use shuffle=3 on video batches? Do I understand right, that shuffle=3 will give me BATCH_SIZe number of random frames from the each video? If it's right, why do we need interval and skip in this mode?

Thanks in advance!

[francoisruty]

I have the same problem:

• shuffle=3 is not working
• I don't understand why we need skip and interval parameters when using an intra-batch shuffle mode (mode 2 or 3)

Right now I want frames to be shuffled, but I want to see all of them, so I specify skip and interval to 0, but all batches returned by VideoLoader are the same