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Added new method get_back_end_interface_set() to speed up back-end in… #5731
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Added new method get_back_end_interface_set() to speed up back-end in…
gechiang b3a9377
Fixed indentation issue introduced when transfeering the changes over…
gechiang 6019bc8
Replaced the block of logic in back-end LAG determination by using on…
gechiang 046e07c
Skip checking LAG if there are no ports in back-end-interf list
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return set(bk_end_intf_list)
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@smaheshm according to some research using .update(xx_list) has better performance when compared to set(xx_list).
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samaddik@samaddik-vm-01:/data/sonic$ python -m timeit 'x = [x for x in range(0, 1000000)]; y = set(); y.update(x)'
10 loops, best of 3: 104 msec per loop
samaddik@samaddik-vm-01:/data/sonic$ python -m timeit 'x = [x for x in range(0, 1000000)]; y = set(); y.update(x)'
10 loops, best of 3: 104 msec per loop
samaddik@samaddik-vm-01:/data/sonic$ python -m timeit 'x = [x for x in range(0, 1000000)]; y = set(); y.update(x)'
10 loops, best of 3: 105 msec per loop
samaddik@samaddik-vm-01:/data/sonic$ python -m timeit 'x = [x for x in range(0, 1000000)]; set(x)'
10 loops, best of 3: 104 msec per loop
samaddik@samaddik-vm-01:/data/sonic$ python -m timeit 'x = [x for x in range(0, 1000000)]; set(x)'
10 loops, best of 3: 105 msec per loop
No noticeable diff for million elements. Either is fine.