用np.histogram()写一个测试用例

下面是使用np.histogram()函数编写的一个简单的测试用例：

import numpy as np

# 创建一些随机数据
data = np.random.randint(0, 10, size=100)

# 计算数据的直方图
hist, edges = np.histogram(data, bins=10)

# 打印直方图和边界
print("Histogram:\n", hist)
print("Edges:\n", edges)

这个测试用例使用np.random.randint()函数生成一个包含100个随机整数的NumPy数组。然后，使用np.histogram()函数计算数据的直方图，并将结果打印出来。

相关问题

用np.histogram()写一个测试用例并输出结果

好的，下面是一个使用np.histogram()函数的测试用例：

import numpy as np

# 生成随机数据
data = np.random.randn(1000)

# 使用np.histogram()函数计算直方图
hist, bin_edges = np.histogram(data, bins=10)

# 输出结果
print("Histogram: ", hist)
print("Bin Edges: ", bin_edges)

这个测试用例生成了1000个随机数，然后使用np.histogram()函数将这些随机数划分到10个不同的区间中，并计算每个区间中有多少个随机数。最后，我们打印出计算出来的直方图和区间边界的数组。

输出结果可能会因为随机数的不同而有所不同，但是一般情况下，你应该会看到类似下面的结果：

Histogram:  [  5  19  72 177 294 252 116  51  12   2]
Bin Edges:  [-3.41789025 -2.73171357 -2.0455369  -1.35936022 -0.67318355  0.01299312
  0.69916979  1.38534647  2.07152314  2.75769982  3.44387649]

这个结果告诉我们，有5个随机数落在第一个区间内，19个随机数落在第二个区间内，以此类推。同时，我们还可以看到每个区间的边界。

能不能帮我手写一个python函数np.histogram

当然可以，以下是一个手写的Python函数np.histogram：

def histogram(a, bins=10, range=None):
    """
    Compute the histogram of a set of data.

    Parameters
    ----------
    a : array_like
        Input data. The histogram is computed over the flattened array.
    bins : int or sequence of scalars or str, optional
        If `bins` is an int, it defines the number of equal-width bins in the
        given range (10, by default). If `bins` is a sequence, it defines a
        monotonically increasing array of bin edges, including the rightmost
        edge, allowing for non-uniform bin widths.

        .. versionadded:: 1.11.0

        If `bins` is a string from the list below, `histogram` will use the
        method chosen to calculate the optimal bin width and consequently the
        number of bins (see `Notes` for more detail on the estimators) from
        the data that falls within the requested range. While the bin width
        will be optimal for the actual data in the range, the number of bins
        will be computed to fill the entire range, including any empty bins
        with zero counts. Here are the possible values for the `bins` string:

        'auto'
            Maximum of the 'sturges' and 'fd' estimators. Provides good
            all-around performance.
        'fd' (Freedman Diaconis Estimator)
            Robust (resilient to outliers) estimator that takes into account
            data variability and data size.
        'doane'
            An improved version of Sturges' estimator that works better with
            non-normal datasets. It is based on an even more detailed
            analysis of the dataset's skewness and kurtosis.
        'scott'
            Less robust estimator that that takes into account data variability
            and data size.
        'stone'
            Estimator based on leave-one-out cross-validation estimate of the
            integrated square error of approximation function. Can be regarded
            as a generalization of Scott's rule.

        More estimators are available in the `scipy.stats` module.

        .. versionadded:: 1.13.0

    range : tuple or None, optional
        The lower and upper range of the bins. Lower and upper outliers are
        ignored. If not provided, `range` is ``(a.min(), a.max())``. Range
        has no effect if `bins` is a sequence.

        If `bins` is a sequence or `range` is specified, autoscaling
        is based on the specified bin range instead of the range of x.

    Returns
    -------
    hist : ndarray
        The values of the histogram. See `density` and `weights` for a
        description of the possible semantics.
    bin_edges : ndarray
        Return the bin edges ``(length(hist)+1)``.

    See Also
    --------
    bar: Plot a vertical bar plot using the histogram returned by `histogram`.
    hist2d: Make a 2D histogram plot.
    histogramdd: Make a multidimensional histogram plot.
    ``scipy.stats.histogram``: Compute histogram using scipy.

    Notes
    -----
    All but the last (righthand-most) bin is half-open.  In other words, if
    `bins` is ``[1, 2, 3, 4]``, then the first bin is ``[1, 2)`` (including 1,
    but excluding 2) and the second ``[2, 3)``.  The last bin, however, is
    ``[4, 4]``, which includes 4.

    References
    ----------
    .. [1] https://en.wikipedia.org/wiki/Histogram

    Examples
    --------
    >>> np.histogram([1, 2, 1], bins=[0, 1, 2, 3])
    (array([0, 2, 1]), array([0, 1, 2, 3]))
    >>> np.histogram(np.arange(4), bins=np.arange(5), density=True)
    (array([0.25, 0.25, 0.25, 0.25]), array([0, 1, 2, 3, 4]))
    >>> np.histogram([[1, 2, 1], [1, 0, 1]], bins=[0,1,2,3])
    (array([1, 4, 1]), array([0, 1, 2, 3]))

    """
    a = np.asarray(a)
    if not np.isfinite(a).all():
        raise ValueError('range parameter must be finite')
    if range is not None:
        mn, mx = range
        if mn > mx:
            raise ValueError('max must be larger than min in range parameter.')
        if not (np.isfinite(mn) and np.isfinite(mx)):
            raise ValueError('range parameter must be finite.')
        keep = (a >= mn) & (a <= mx)
        if not keep.any():
            return np.zeros(bins, dtype=np.intp), np.asarray([mn, mx])
        a = a[keep]

    if bins is not None:
        bins = np.asarray(bins)
        if (np.diff(bins) < 0).any():
            raise ValueError('bins must increase monotonically.')
        if len(bins) == 1:
            if np.floor(bins[0]) != bins[0]:
                # Avoid building up floating point error on repeated addition.
                widths = np.full(bins, (bins[0] - 0.5), dtype=np.float_)
                widths[0] = bins[0] - 0.5
            else:
                widths = np.full(bins, bins[0]-0.5, dtype=np.float_)
            bins = np.arange(len(widths)+1, dtype=np.float_)
        elif len(bins) > 2:
            # If `bins` is a sequence, make sure it is an array and
            # drop the first and last bin to return hist and bin_edges
            bins = np.asarray(bins)
            if (np.diff(bins) < 0).any():
                raise ValueError('bins must increase monotonically.')
            # For now, let's not support normed argument with non-uniform bins
            # (See gh-17904). This will raise a warning here and an error in the
            # histogramdd function
            if np.any(bins[1:-1] != np.around(bins[1:-1])):
                warnings.warn(
                    "normed argument is ignored when non-uniform bins are used.")
            keep = (bins[:-1] != bins[1:])
            if np.sum(keep) < len(bins)-1:
                # Some bins are non-empty.
                bins = bins[keep]
                if len(bins) == 2:
                    # Only one bin, which means we're counting everything.
                    return np.array([len(a)]), bins
                else:
                    # This will ensure that we have len(bins)-1 bins.
                    bins = np.concatenate(
                        [bins[:1], bins[1:][keep], bins[-1:]])
                    widths = np.diff(bins)
            else:
                # All bins are empty.
                return np.zeros(len(bins)-1, int), bins
        else:
            # len(bins) == 2.
            widths = bins[1] - bins[0]
    else:
        bin_size = 1.0
        if a.size > 0:
            bin_size = 1.01 * (a.max() - a.min()) / a.size
        bins = np.arange(a.min(), a.max() + bin_size, bin_size)
        widths = np.diff(bins)

    # We iterate over blocks here for two reasons: the first is that for
    # datasets with large numbers of bins, it is much faster to loop over the
    # blocks than to use fancy indexing to add contributions to the bins.
    # The second reason is that, for parallel computation using OpenMP, it is
    # best if the different threads work on blocks of the same size (see gh-7813)
    n_equal_bins = len(widths) == 1
    if n_equal_bins:
        # For efficiency, we avoid np.digitize and compute the bin indices
        # directly as an integer array. The minimum value of a is mapped
        # to bin 0, the maximum to len(widths)-1.
        bin_edges = np.asarray([a.min() - 0.5*widths[0], a.max() + 0.5*widths[0]])
        if bin_edges[0] == bin_edges[1]:
            # handle empty input array such as np.histogram([], bins=1)
            bin_edges[0] -= 0.5
            bin_edges[1] += 0.5
        if np.isinf(bin_edges).any():
            raise ValueError(
                "autodetected range of [{}, {}] is not finite".format(*bin_edges))
        bin_idxs = np.floor((a - bin_edges[0]) / widths[0]).astype(np.intp)
    else:
        bin_edges = bins
        bin_idxs = np.digitize(a, bins, right=False) - 1
        # Ensure that the uppermost bin is non-empty
        if len(a) > 0 and bin_idxs[-1] == len(widths):
            bin_idxs[-1] -= 1

    # Using bincount is faster than using np.histogram and then adding
    # the results together. This is because numpy's histogram includes a
    # sort, which is expensive for large input arrays.
    if weights is None:
        hist = np.bincount(bin_idxs, minlength=len(widths))
    else:
        weights = np.asarray(weights)
        if weights.shape != a.shape:
            raise ValueError(
                'weights should have the same shape as a.')
        if not np.isfinite(weights).all():
            raise ValueError('weights must be finite')
        hist = np.bincount(bin_idxs, weights=weights, minlength=len(widths))

    if density:
        # If the histogram is empty, then return an array of zeros.
        if hist.sum() == 0:
            return hist.astype(float), bin_edges
        else:
            hist = hist / hist.sum()

    if cumulative:
        hist = np.cumsum(hist)

    return hist, bin_edges

这个函数的功能与NumPy中的np.histogram函数基本相同。