在标准正态分布表中查找值

Question

我正在编写一个代码，从附表中给定一个值z。该表的读取方式是：给定值z=1.34，可以从条目1.3（行的第一个值）和0.04（第一行的值）中找到z的值。在本例中，x（1.34）=0.4099。我试图编写一个代码，输出给定z（本例中为1.34）的z值（本例中为0.499）

我编写了以下代码：

def standard_noraml_z(z1):
    standard_normal=[[0.00, 0.01, 0.02, 0.03, 0.04, 0.05, 0.06, 0.07, 0.08, 0.09],
    [0.0, 0.0000, 0.0040, 0.0080, 0.0120, 0.0160, 0.0199, 0.0239, 0.0279, 0.0319, 0.0359],
    [0.1, 0.0398, 0.0438, 0.0478, 0.0517, 0.0557, 0.0596, 0.0636, 0.0675, 0.0714, 0.0753],
    [0.2, 0.0793, 0.0832, 0.0871, 0.0910, 0.0948, 0.0987, 0.1026, 0.1064, 0.1103, 0.1141],
    [0.3, 0.1179, 0.1217, 0.1255, 0.1293, 0.1331, 0.1368, 0.1406, 0.1443, 0.1480, 0.1517],
    [0.4, 0.1554, 0.1591, 0.1628, 0.1664, 0.1700, 0.1736, 0.1772, 0.1809, 0.1844, 0.1879],
    [0.5, 0.1915, 0.1950, 0.1985, 0.2019, 0.2054, 0.2088, 0.2123, 0.2157, 0.2190, 0.2224],
    [0.6, 0.2257, 0.2291, 0.2324, 0.2357, 0.2389, 0.2422, 0.2454, 0.2486, 0.2517, 0.2549],
    [0.7, 0.2580, 0.2611, 0.2642, 0.2673, 0.2704, 0.2734, 0.2764, 0.2794, 0.2823, 0.2852],
    [0.8, 0.2881, 0.2910, 0.2939, 0.2967, 0.2995, 0.3023, 0.3051, 0.3078, 0.3106, 0.3133],
    [0.9, 0.3159, 0.3186, 0.3212, 0.3238, 0.3264, 0.3289, 0.3315, 0.3340, 0.3365, 0.3389],
    [1.0, 0.3413, 0.3438, 0.3461, 0.3485, 0.3508, 0.3531, 0.3554, 0.3577, 0.3599, 0.3621],
    [1.1, 0.3643, 0.3665, 0.3686, 0.3708, 0.3729, 0.3749, 0.3770, 0.3790, 0.3810, 0.3830],
    [1.2, 0.3849, 0.3869, 0.3888, 0.3907, 0.3925, 0.3944, 0.3962, 0.3980, 0.3997, 0.4015],
    [1.3, 0.4032, 0.4049, 0.4066, 0.4082, 0.4099, 0.4115, 0.4131, 0.4147, 0.4162, 0.4177],
    [1.4, 0.4192, 0.4207, 0.4222, 0.4236, 0.4251, 0.4265, 0.4279, 0.4292, 0.4306, 0.4319],
    [1.5, 0.4332, 0.4345, 0.4357, 0.4370, 0.4382, 0.4394, 0.4406, 0.4418, 0.4429, 0.4441],
    [1.6, 0.4452, 0.4463, 0.4474, 0.4484, 0.4495, 0.4505, 0.4515, 0.4525, 0.4535, 0.4545],
    [1.7, 0.4554, 0.4564, 0.4573, 0.4582, 0.4591, 0.4599, 0.4608, 0.4616, 0.4625, 0.4633],
    [1.8, 0.4641, 0.4649, 0.4656, 0.4664, 0.4671, 0.4678, 0.4686, 0.4693, 0.4699, 0.4706],
    [1.9, 0.4713, 0.4719, 0.4726, 0.4732 , 0.4738, 0.4744, 0.4750, 0.4758, 0.4761, 0.4767],
    [2.0, 0.4772, 0.4778, 0.4783, 0.4788, 0.4793, 0.4799, 0.4803, 0.4808, 0.4812, 0.4817],
    [2.1, 0.4821, 0.4826, 0.4830, 0.4834, 0.4838, 0.4842, 0.4846, 0.4850, 0.4854, 0.4857],
    [2.2, 0.4861, 0.4864, 0.4868, 0.4871, 0.4875, 0.4878, 0.4881, 0.4884, 0.4887, 0.4890],
    [2.3, 0.4893, 0.4896, 0.4898, 0.4901, 0.4904, 0.4906, 0.4909, 0.4911, 0.4913, 0.4916],
    [2.4, 0.4918, 0.4920, 0.4922, 0.4925, 0.4927, 0.4929, 0.4931, 0.4932, 0.4934, 0.4936],
    [2.5, 0.4938, 0.4940, 0.4941, 0.4943, 0.4945, 0.4946, 0.4948, 0.4949, 0.4951, 0.4952],
    [2.6, 0.4953, 0.4955, 0.4956, 0.4957, 0.4959, 0.4960, 0.4961, 0.4962, 0.4963, 0.4964],
    [2.7, 0.4965, 0.4966, 0.4967, 0.4968, 0.4969, 0.4970, 0.4971, 0.4972, 0.4973, 0.4974],
    [2.8, 0.4974, 0.4975, 0.4976, 0.4977, 0.4977, 0.4978, 0.4979, 0.4979, 0.4980, 0.4981],
    [2.9, 0.4981, 0.4982, 0.4982, 0.4983, 0.4984, 0.4984, 0.4985, 0.4985, 0.4986, 0.4986],
    [3.0, 0.49865, 0.4987, 0.4987, 0.4988, 0.4988, 0.4988, 0.4989, 0.4989, 0.4989, 0.4990]]
    z=str(z1)
    zy=z[1:3]
    zx=z[4]
    y=0
    x=0
    for i in range(len(standard_normal)):
        if standard_normal[i][0]==zy:
            y=i
    for i in range(len(standard_normal[0])):
        if standard_normal[0][i]==zx:
            x=i+1
    return standard_normal[y][x]

print(standard_noraml_z(1.34))

Answer 1

当前代码返回错误，因为未使用零索引。要从字符串中提取前三个值，可以写z[0:3]或只写z[:3]。然后，要获得第四个值，您需要写入z[3]。但是，由于实际上需要将第四个值从4转换为0.04，并且需要将其作为数值，而不是字符串，因此只需使用数学函数来获取zx和zy值就更简单了

因此，替换：

    z=str(z1)
    zy=z[1:3]
    zx=z[4]

与：

    import math

    zy = math.floor(z1 * 10)/10.0
    zx = round(z1 - zy,2)