Hi Neal, Le 14/01/2013 15:39, Neal Becker a écrit : > This code should explain all: > -------------------------------- > import numpy as np > arg = np.angle > > def nint (x): > return int (x + 0.5) if x >= 0 else int (x - 0.5) > > def unwrap (inp, y=np.pi, init=0, cnt=0): > o = np.empty_like (inp) > prev_o = init > for i in range (len (inp)): > o[i] = cnt * 2 * y + inp[i] > delta = o[i] - prev_o > > if delta / y > 1 or delta / y < -1: > n = nint (delta / (2*y)) > o[i] -= 2*y*n > cnt -= n > > prev_o = o[i] > > return o > > > u = np.linspace (0, 400, 100) * np.pi/100 > v = np.cos (u) + 1j * np.sin (u) > plot (arg(v)) > plot (arg(v) + arg (v)) > plot (unwrap (arg (v))) > plot (unwrap (arg (v) + arg (v))) I think your code does the job.
I tried the following simplification, without the use of nint (which by
the way could be replaced by int(floor(x)) I think) :
def unwrap (inp, y=np.pi, init=0, cnt=0):
o = np.empty_like (inp)
prev_o = init
for i in range (len (inp)):
o[i] = cnt * 2 * y + inp[i]
delta = o[i] - prev_o
if delta / y > 1:
o[i] -= 2*y
cnt -= 1
elif delta / y < -1:
o[i] += 2*y
cnt += 1
prev_o = o[i]
return o
And now I understand the issue you described of "phase changes of more
than 2pi" because the above indeed fail to unwrap (arg (v) + arg (v)).
On the other hand np.unwrap handles it correctly.
(I still don't know for the speed issue).
Best,
Pierre
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