python查看函数源代码

除了在系统中找到源文件直接打开查看，还可以用inspect.getsource函数查看

示例1

import inspectdef print_hh():print('hh')source = inspect.getsource(print_hh) # 查看自定义函数print_hh的源代码
print(source)

运行结果：

def print_hh():print('hh')

示例2

import inspect
import numpy as npsource = inspect.getsource(np.sum) # 查看系统函数numpy加和函数sum的源代码
print(source)

运行结果：

@array_function_dispatch(_sum_dispatcher)
def sum(a, axis=None, dtype=None, out=None, keepdims=np._NoValue,initial=np._NoValue, where=np._NoValue):"""Sum of array elements over a given axis.Parameters----------a : array_likeElements to sum.axis : None or int or tuple of ints, optionalAxis or axes along which a sum is performed.  The default,axis=None, will sum all of the elements of the input array.  Ifaxis is negative it counts from the last to the first axis... versionadded:: 1.7.0If axis is a tuple of ints, a sum is performed on all of the axesspecified in the tuple instead of a single axis or all the axes asbefore.dtype : dtype, optionalThe type of the returned array and of the accumulator in which theelements are summed.  The dtype of `a` is used by default unless `a`has an integer dtype of less precision than the default platforminteger.  In that case, if `a` is signed then the platform integeris used while if `a` is unsigned then an unsigned integer of thesame precision as the platform integer is used.out : ndarray, optionalAlternative output array in which to place the result. It must havethe same shape as the expected output, but the type of the outputvalues will be cast if necessary.keepdims : bool, optionalIf this is set to True, the axes which are reduced are leftin the result as dimensions with size one. With this option,the result will broadcast correctly against the input array.If the default value is passed, then `keepdims` will not bepassed through to the `sum` method of sub-classes of`ndarray`, however any non-default value will be.  If thesub-class' method does not implement `keepdims` anyexceptions will be raised.initial : scalar, optionalStarting value for the sum. See `~numpy.ufunc.reduce` for details... versionadded:: 1.15.0where : array_like of bool, optionalElements to include in the sum. See `~numpy.ufunc.reduce` for details... versionadded:: 1.17.0Returns-------sum_along_axis : ndarrayAn array with the same shape as `a`, with the specifiedaxis removed.   If `a` is a 0-d array, or if `axis` is None, a scalaris returned.  If an output array is specified, a reference to`out` is returned.See Also--------ndarray.sum : Equivalent method.add.reduce : Equivalent functionality of `add`.cumsum : Cumulative sum of array elements.trapz : Integration of array values using the composite trapezoidal rule.mean, averageNotes-----Arithmetic is modular when using integer types, and no error israised on overflow.The sum of an empty array is the neutral element 0:>>> np.sum([])0.0For floating point numbers the numerical precision of sum (and``np.add.reduce``) is in general limited by directly adding each numberindividually to the result causing rounding errors in every step.However, often numpy will use a  numerically better approach (partialpairwise summation) leading to improved precision in many use-cases.This improved precision is always provided when no ``axis`` is given.When ``axis`` is given, it will depend on which axis is summed.Technically, to provide the best speed possible, the improved precisionis only used when the summation is along the fast axis in memory.Note that the exact precision may vary depending on other parameters.In contrast to NumPy, Python's ``math.fsum`` function uses a slower butmore precise approach to summation.Especially when summing a large number of lower precision floating pointnumbers, such as ``float32``, numerical errors can become significant.In such cases it can be advisable to use `dtype="float64"` to use a higherprecision for the output.Examples-------->>> np.sum([0.5, 1.5])2.0>>> np.sum([0.5, 0.7, 0.2, 1.5], dtype=np.int32)1>>> np.sum([[0, 1], [0, 5]])6>>> np.sum([[0, 1], [0, 5]], axis=0)array([0, 6])>>> np.sum([[0, 1], [0, 5]], axis=1)array([1, 5])>>> np.sum([[0, 1], [np.nan, 5]], where=[False, True], axis=1)array([1., 5.])If the accumulator is too small, overflow occurs:>>> np.ones(128, dtype=np.int8).sum(dtype=np.int8)-128You can also start the sum with a value other than zero:>>> np.sum([10], initial=5)15"""if isinstance(a, _gentype):# 2018-02-25, 1.15.0warnings.warn("Calling np.sum(generator) is deprecated, and in the future will give a different result. ""Use np.sum(np.fromiter(generator)) or the python sum builtin instead.",DeprecationWarning, stacklevel=3)res = _sum_(a)if out is not None:out[...] = resreturn outreturn resreturn _wrapreduction(a, np.add, 'sum', axis, dtype, out, keepdims=keepdims,initial=initial, where=where)

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python查看函数源代码

python查看函数源代码

示例1

示例2

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