def resize(self, first_day, last_day):
'''Return a copy of myself with new bounds first_day and last_day.
>>> a = Date_Vector('2013-06-02', np.arange(2, 7))
>>> a
Date_Vector('2013-06-02', [2, 3, 4, 5, 6])
Shrinking removes the newly extra elements (note that if either bound
is None, we use the existing bound):
>>> s = a.resize('2013-06-03', None)
>>> s
Date_Vector('2013-06-03', [3, 4, 5, 6])
>>> a.resize(None, '2013-06-04')
Date_Vector('2013-06-02', [2, 3, 4])
>>> a.resize('2013-06-03', '2013-06-04')
Date_Vector('2013-06-03', [3, 4])
>>> a.resize('2013-06-06', None)
Date_Vector('2013-06-06', [6])
>>> a.resize(None, '2013-06-02')
Date_Vector('2013-06-02', [2])
If the new bounds cross, return None:
>>> a.resize('2013-06-07', None) is None
True
>>> a.resize(None, '2013-06-01') is None
True
If shrinking, the result may (but is not guaranteed to) be a view of
the original vector:
>>> np.may_share_memory(a, s)
True
Growing adds new elements containing zero:
>>> g = a.resize('2013-06-01', '2013-06-07')
>>> g
Date_Vector('2013-06-01', [0, 2, 3, 4, 5, 6, 0])
>>> a.resize('2013-06-01', None)
Date_Vector('2013-06-01', [0, 2, 3, 4, 5, 6])
>>> a.resize(None, '2013-06-07')
Date_Vector('2013-06-02', [2, 3, 4, 5, 6, 0])
You can grow and shrink at the same time:
>>> gs = a.resize('2013-06-01', '2013-06-05')
>>> gs
Date_Vector('2013-06-01', [0, 2, 3, 4, 5])
>>> a.resize('2013-06-03', '2013-06-07')
Date_Vector('2013-06-03', [3, 4, 5, 6, 0])
Grown vectors are not views:
>>> np.may_share_memory(a, g)
False
>>> np.may_share_memory(a, gs)
False
Finally, it's OK to do a no-op resize. In this case, the result is a
shallow copy.
>>> n = a.resize(None, None)
>>> n
Date_Vector('2013-06-02', [2, 3, 4, 5, 6])
>>> n is a
False
>>> np.may_share_memory(a, n)
True'''
# clean up new bounds
fd_new = time_.dateify(first_day) or self.first_day
ld_new = time_.dateify(last_day) or self.last_day
# if they're empty, return None
if (max(fd_new, self.first_day) > min(ld_new, self.last_day)):
return None
# how many elements to add and remove from the start?
delta_start = time_.days_diff(fd_new, self.first_day)
trim_start = max(0, delta_start)
add_start = max(0, -delta_start)
# how many elements to add and remove from the end?
delta_end = time_.days_diff(self.last_day, ld_new)
trim_end = max(0, delta_end)
add_end = max(0, -delta_end)
# Do it!
if (add_start == 0 and add_end == 0):
# If shrinking, don't use hstack(); this avoids copying data, which
# favors speed over memory. The caller can do a deep copy if this is
# a problem.
return Date_Vector(fd_new, self[trim_start:len(self) - trim_end])
else:
return Date_Vector(fd_new,
np.hstack([np.zeros(add_start, dtype=self.dtype),
self[trim_start:len(self) - trim_end],
np.zeros(add_end, dtype=self.dtype)]))
def resize(self, first_day, last_day):
'''Return a copy of myself with new bounds first_day and last_day.
>>> a = Date_Vector('2013-06-02', np.arange(2, 7))
>>> a
Date_Vector('2013-06-02', [2, 3, 4, 5, 6])
Shrinking removes the newly extra elements (note that if either bound
is None, we use the existing bound):
>>> s = a.resize('2013-06-03', None)
>>> s
Date_Vector('2013-06-03', [3, 4, 5, 6])
>>> a.resize(None, '2013-06-04')
Date_Vector('2013-06-02', [2, 3, 4])
>>> a.resize('2013-06-03', '2013-06-04')
Date_Vector('2013-06-03', [3, 4])
>>> a.resize('2013-06-06', None)
Date_Vector('2013-06-06', [6])
>>> a.resize(None, '2013-06-02')
Date_Vector('2013-06-02', [2])
If the new bounds cross, return None:
>>> a.resize('2013-06-07', None) is None
True
>>> a.resize(None, '2013-06-01') is None
True
If shrinking, the result may (but is not guaranteed to) be a view of
the original vector:
>>> np.may_share_memory(a, s)
True
Growing adds new elements containing zero:
>>> g = a.resize('2013-06-01', '2013-06-07')
>>> g
Date_Vector('2013-06-01', [0, 2, 3, 4, 5, 6, 0])
>>> a.resize('2013-06-01', None)
Date_Vector('2013-06-01', [0, 2, 3, 4, 5, 6])
>>> a.resize(None, '2013-06-07')
Date_Vector('2013-06-02', [2, 3, 4, 5, 6, 0])
You can grow and shrink at the same time:
>>> gs = a.resize('2013-06-01', '2013-06-05')
>>> gs
Date_Vector('2013-06-01', [0, 2, 3, 4, 5])
>>> a.resize('2013-06-03', '2013-06-07')
Date_Vector('2013-06-03', [3, 4, 5, 6, 0])
Grown vectors are not views:
>>> np.may_share_memory(a, g)
False
>>> np.may_share_memory(a, gs)
False
Finally, it's OK to do a no-op resize. In this case, the result is a
shallow copy.
>>> n = a.resize(None, None)
>>> n
Date_Vector('2013-06-02', [2, 3, 4, 5, 6])
>>> n is a
False
>>> np.may_share_memory(a, n)
True'''
# clean up new bounds
fd_new = time_.dateify(first_day) or self.first_day
ld_new = time_.dateify(last_day) or self.last_day
# if they're empty, return None
if (max(fd_new, self.first_day) > min(ld_new, self.last_day)):
return None
# how many elements to add and remove from the start?
delta_start = time_.days_diff(fd_new, self.first_day)
trim_start = max(0, delta_start)
add_start = max(0, -delta_start)
# how many elements to add and remove from the end?
delta_end = time_.days_diff(self.last_day, ld_new)
trim_end = max(0, delta_end)
add_end = max(0, -delta_end)
# Do it!
if (add_start == 0 and add_end == 0):
# If shrinking, don't use hstack(); this avoids copying data, which
# favors speed over memory. The caller can do a deep copy if this is
# a problem.
return Date_Vector(fd_new, self[trim_start:len(self) - trim_end])
else:
return Date_Vector(
fd_new,
np.hstack([
np.zeros(add_start, dtype=self.dtype),
self[trim_start:len(self) - trim_end],
np.zeros(add_end, dtype=self.dtype)
]))
def first_day(self, x):
self._first_day = time_.dateify(x)
def first_day(self, x):
self._first_day = time_.dateify(x)
