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98 lines
2.4 KiB
Python
98 lines
2.4 KiB
Python
#
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# Copyright 2013 Quantopian, Inc.
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#
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# Licensed under the Apache License, Version 2.0 (the "License");
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# you may not use this file except in compliance with the License.
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# You may obtain a copy of the License at
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#
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# http://www.apache.org/licenses/LICENSE-2.0
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#
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# Unless required by applicable law or agreed to in writing, software
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# distributed under the License is distributed on an "AS IS" BASIS,
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# WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
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# See the License for the specific language governing permissions and
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# limitations under the License.
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from decimal import Decimal
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import math
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from numpy import isnan
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def tolerant_equals(a, b, atol=10e-7, rtol=10e-7, equal_nan=False):
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"""Check if a and b are equal with some tolerance.
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Parameters
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----------
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a, b : float
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The floats to check for equality.
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atol : float, optional
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The absolute tolerance.
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rtol : float, optional
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The relative tolerance.
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equal_nan : bool, optional
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Should NaN compare equal?
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See Also
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--------
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numpy.isclose
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Notes
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-----
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This function is just a scalar version of numpy.isclose for performance.
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See the docstring of ``isclose`` for more information about ``atol`` and
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``rtol``.
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"""
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if equal_nan and isnan(a) and isnan(b):
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return True
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return math.fabs(a - b) <= (atol + rtol * math.fabs(b))
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try:
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# fast versions
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import bottleneck as bn
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nanmean = bn.nanmean
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nanstd = bn.nanstd
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nansum = bn.nansum
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nanmax = bn.nanmax
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nanmin = bn.nanmin
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nanargmax = bn.nanargmax
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nanargmin = bn.nanargmin
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except ImportError:
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# slower numpy
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import numpy as np
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nanmean = np.nanmean
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nanstd = np.nanstd
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nansum = np.nansum
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nanmax = np.nanmax
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nanmin = np.nanmin
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nanargmax = np.nanargmax
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nanargmin = np.nanargmin
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def round_if_near_integer(a, epsilon=1e-4):
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"""
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Round a to the nearest integer if that integer is within an epsilon
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of a.
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"""
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if abs(a - round(a)) <= epsilon:
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return round(a)
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else:
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return a
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def number_of_decimal_places(n):
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"""
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Compute the number of decimal places in a number.
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Examples
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--------
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>>> number_of_decimal_places(1)
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0
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>>> number_of_decimal_places(3.14)
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2
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>>> number_of_decimal_places('3.14')
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2
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"""
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decimal = Decimal(str(n))
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return -decimal.as_tuple().exponent
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