1

我想在 jupyter (4.0.6) 笔记本中使用 matplotlib (1.5.1) 进行交互式绘图。问题是静态图是使用具有四个变量的函数创建的,其中两个是常量,其中两个是关键字参数,我想以交互方式更改关键字参数。

这是可能的,如果是的话,怎么做?

下面的概念代码显示了生成绘图的函数make_figure(...)和生成交互式绘图的命令。

如果我将关键字参数更改为变量,则会收到错误消息“interact() 采用 0 到 1 个位置参数,但给出了 3 个”

概念代码:

def make_figure(const_1, const_2, var_1=0.4, var_2=0.8):
    b = calc_b(var_1, var_2)
    c = calc_c(b, const_1, const_2)
    fig, ax = plt.subplots()
    N, bins, patches = ax.hist(c)


interact(make_figure, 
         const_1,
         const_2,
         var_1=(0.2, 0.4, 0.05),
         var_2=(0.75, 0.95, 0.05))

添加 20160325:代码示例

我正在尝试为一个班级的分数创建一个直方图,这取决于分别达到 1.0 和 4.0 所需的百分比。

# setup some marks
ids_perc = np.random.random(33) 
print("number of entered marks: ", ids_perc.shape)

直方图的主要代码;主功能:get_marks

# define possible marks
marks = np.array([1.0,
                  1.3,
                  1.7,
                  2.0,
                  2.3,
                  2.7,
                  3.0,
                  3.3,
                  3.7,
                  4.0,
                  5.0])
marks_possible = marks[::-1]

def get_perc_necessary(min_perc_one,
                       min_perc_four,
                       n_marks):
    """
    calculates an equally spaced array for percentage necessary to get a mark
    """
    delta = (min_perc_one - min_perc_four)/(n_marks-2-1)
    perc_necessary_raw = np.linspace(start=min_perc_four, 
                                     stop=min_perc_one, 
                                     num=n_marks-1)
    perc_necessary = np.append([0.0], np.round(perc_necessary_raw, decimals=2)) 
    return perc_necessary


def assign_marks(n_students,
                 perc_necessary,
                 achieved_perc,
                 marks_real):
    """
    get the mark for each student (with a certain achieved percentage)
    """
    final_marks = np.empty(n_students)

    for cur_i in range(n_students):
        idx = np.argmax(np.argwhere(perc_necessary <= achieved_perc[cur_i]))
        final_marks[cur_i] = marks_real[idx]

    return final_marks


def get_marks(achieved_perc = ids_perc,
              marks_real = marks_possible,                    
              min_perc_four = 0.15,
              min_perc_one = 0.85):

    n_marks = marks.shape[0]
#     print("n_marks: ", n_marks)
    n_students = achieved_perc.shape[0]
#     print("n_students: ", n_students)

    # -----------------------------
    # linear step between each mark
    perc_necessary = get_perc_necessary(min_perc_one,
                                        min_perc_four,
                                        n_marks)

    # test query: there need to be as many percentages as marks
    if perc_necessary.shape[0] != marks_real.shape[0]:
        print("the number of marks has to be equal the number of boundaries")
        raise Exception

    # ------------
    # assign marks 
    final_marks = assign_marks(n_students,
                               perc_necessary,
                               achieved_perc,
                               marks_real)    

    # ------------
    # create table
    fig, ax = plt.subplots()
    N, bins, patches = ax.hist(final_marks, 
                               align='mid', 
                               bins=np.append(marks,6.)) # bins=marks
    ax.xaxis.set_major_formatter(FormatStrFormatter('%0.1f'))
    bin_centers = 0.5 * np.diff(bins) + bins[:-1]
    ax.set_xticks(bin_centers)
    ax.set_xticklabels( marks )
    ax.set_xlabel("mark")
    ax.set_ylabel("number of marks")
    ax.set_ylim(0.0, 6.0)
    plt.grid(True)

现在,当我尝试设置interact这样做时

interact(get_marks, 
     min_perc_four=(0.2, 0.4, 0.05),
     min_perc_one=(0.75, 0.95, 0.05));

我得到错误

ValueError: array([ 0.22366653,  0.74206953,  0.47501716,  0.56536227,  0.54792759,
    0.60288287,  0.68548973,  0.576935  ,  0.84582243,  0.40709693,
    0.78600622,  0.2692508 ,  0.62524819,  0.62204851,  0.5421716 ,
    0.71836192,  0.97194698,  0.4054752 ,  0.2185643 ,  0.11786751,
    0.57947848,  0.88659768,  0.38803576,  0.66617254,  0.77663263,
    0.94364543,  0.23021637,  0.30899724,  0.08695842,  0.50296694,
    0.8164095 ,  0.77892531,  0.5542163 ]) cannot be transformed to a Widget

为什么查看变量时出现此错误ids_perc

4

1 回答 1

1

您需要在interact(). 例如,像这样:

const_1 = 1

interact(make_figure, 
         const_1=const_1,
         const_2=2,
         var_1=(0.2, 0.4, 0.05),
         var_2=(0.75, 0.95, 0.05))

或者(如果可能)更改签名make_figure以使这些变量成为关键字参数,这样您就可以避免显式传递它们:

def make_figure(const_1=1, const_2=2, var_1=0.4, var_2=0.8):
    ....

interact(make_figure, 
         var_1=(0.2, 0.4, 0.05),
         var_2=(0.75, 0.95, 0.05))

这是您可以尝试的 MCWE:

def calc_b(v1, v2):
    return v1 + v2

def calc_c(v1, v2, v3):
    return [v1, v2, v3]

def make_figure(const_1=1, const_2=2, var_1=0.4, var_2=0.8):
    b = calc_b(var_1, var_2)
    c = calc_c(b, const_1, const_2)
    fig, ax = plt.subplots()
    N, bins, patches = ax.hist(c)

interact(make_figure, 
         var_1=(0.2, 0.4, 0.05),
         var_2=(0.75, 0.95, 0.05));

这运行没有任何错误。

在你的addition 20160325

您传递给交互的每个参数都必须可以由以下之一表示(稍微简化一下):

  • 一个滑块(对于tuples,表示(最小值,最大值)和标量数)
  • 选择框(用于字符串和字典列表)
  • 复选框(用于布尔值)
  • 输入框(用于字符串)

您正在传递(通过在您的get_marks两个参数中隐式定义为np.arrays)。所以interact不知道如何在滑块上表示它,因此错误。

你至少有两个选择:

1)更改的签名,get_marks以便它采用interact将无法识别的参数(参见上面的项目符号列表)

2)制作另一个包装函数,该函数将采用interact未定义的参数,但将get_marks 将这些参数转换为任何get_marks需要后调用。

因此,只需多走一步,您就完成了。;-)

更新

这是对我有用的带有包装器的代码。请注意,get_marks_interact不需要获取所有参数get_marks并且我不传递列表,因为interact它们会有问题(列表应该表示字符串列表(对于下拉小部件)或[min, max]值列表/元组(对于滑块)) .

def get_marks(min_perc_four = 0.15,
              min_perc_one = 0.85,
              marks=marks_possible,
              ach_per=ids_perc):

    marks_real = marks #  [0]
    achieved_perc = ach_per #  [0]

    n_marks = marks_real.shape[0]
    print("n_marks: ", n_marks)
    n_students = achieved_perc.shape[0]
    print("n_students: ", n_students)

    # -----------------------------
    # linear step between each mark
    perc_necessary = get_perc_necessary(min_perc_one,
                                        min_perc_four,
                                        n_marks)

    # test query: there need to be as many percentages as marks
    if perc_necessary.shape[0] != marks_real.shape[0]:
        print("the number of marks has to be equal the number of boundaries")
        raise Exception

    # ------------
    # assign marks 
    final_marks = assign_marks(n_students,
                               perc_necessary,
                               achieved_perc,
                               marks_real)

    # ------------
    # create table
    fig, ax = plt.subplots()
    N, bins, patches = ax.hist(final_marks, 
                               align='mid', 
                               bins=np.sort(np.append(marks, 6.))) # bins=marks
    ax.xaxis.set_major_formatter(FormatStrFormatter('%0.1f'))
    bin_centers = 0.5 * np.diff(bins) + bins[:-1]
    ax.set_xticks(bin_centers)
    ax.set_xticklabels( marks )
    ax.set_xlabel("mark")
    ax.set_ylabel("number of marks")
    ax.set_ylim(0.0, 6.0)
    plt.grid(True)

def get_marks_interact(min_perc_four = 0.15, 
                       min_perc_one = 0.85,):
    return get_marks(min_perc_four, min_perc_one)

interact(get_marks_wrapper, 
         min_perc_four=(0.2, 0.4, 0.05),
         min_perc_one=(0.75, 0.95, 0.05));
于 2016-03-24T12:44:56.657 回答