Question

无论我使用matplotlib，Open-Flash-Charts还是其他图表框架，我总是需要找到一种方法来设置x / y比例限制和间隔，因为内置的智能不够（或根本没有... ）

只需在pylab（ipyhton -pylab）中尝试一下，了解我的意思：

In [1]: a, b, x = np.zeros(10), np.ones(10), np.arange(10)

In [2]: plot(x, a); plot(x, b)

你会看到正好和空的框架网格隐藏了它的顶部和底部边框下面的2条水平线。

我想知道是否有一些算法（我可以移植到python）来巧妙地设置顶部和底部y限制和步骤，并计算每个显示x厚度的值。

例如，假设我有(datetime, temperature)的475个度量为(x, y)

2011-01-15 10:45:00 < datetime < 2011-01-17 02:20:00

（每5分钟一次）和

26.5 < temperature < 28.3

我对这个特殊情况的建议可能是：

26.4 <= y_scale <= 28.4，每.2

每隔12项（每小时一次）

并在x_scale上打勾。

但是，如果我在-21.5 < temperature < 38.7的20天内只有20项措施，那么呢？是否有标准化的方法？

Answer 1

以下是我多年来一直使用的简单而且运作良好的内容。请原谅我是C，但转换为Python并不困难。

需要以下功能，来自Graphic Gems第1卷。

double NiceNumber (const double Value, const int Round) {
  int    Exponent;
  double Fraction;
  double NiceFraction;

  Exponent = (int) floor(log10(Value));
  Fraction = Value/pow(10, (double)Exponent);

  if (Round) {
    if (Fraction < 1.5) 
      NiceFraction = 1.0;
    else if (Fraction < 3.0)
      NiceFraction = 2.0;
    else if (Fraction < 7.0)
      NiceFraction = 5.0;
    else
      NiceFraction = 10.0;
   }
  else {
    if (Fraction <= 1.0)
      NiceFraction = 1.0;
    else if (Fraction <= 2.0)
      NiceFraction = 2.0;
    else if (Fraction <= 5.0)
      NiceFraction = 5.0;
    else
      NiceFraction = 10.0;
   }

  return NiceFraction*pow(10, (double)Exponent);
 }

使用它，如下例所示，根据您希望显示的主刻度数选择轴的“漂亮”开始/结束。如果您不关心刻度线，可以将其设置为常数值（例如：10）。

      //Input parameters
  double AxisStart = 26.5;
  double AxisEnd   = 28.3;
  double NumTicks  = 10;

  double AxisWidth;
  double NewAxisStart;
  double NewAxisEnd;
  double NiceRange;
  double NiceTick;

    /* Check for special cases */
  AxisWidth = AxisEnd - AxisStart;
  if (AxisWidth == 0.0) return (0.0);

    /* Compute the new nice range and ticks */
  NiceRange = NiceNumber(AxisEnd - AxisStart, 0);
  NiceTick = NiceNumber(NiceRange/(NumTicks - 1), 1);

    /* Compute the new nice start and end values */
  NewAxisStart = floor(AxisStart/NiceTick)*NiceTick;
  NewAxisEnd = ceil(AxisEnd/NiceTick)*NiceTick;

  AxisStart = NewAxisStart; //26.4
  AxisEnd = NewAxisEnd;     //28.4

Answer 2

我在这里报告我的上述C代码的python版本，如果它可能对某人有任何帮助：

import math

def nice_number(value, round_=False):
    '''nice_number(value, round_=False) -> float'''
    exponent = math.floor(math.log(value, 10))
    fraction = value / 10 ** exponent

    if round_:
        if fraction < 1.5: nice_fraction = 1.
        elif fraction < 3.: nice_fraction = 2.
        elif fraction < 7.: nice_fraction = 5.
        else: niceFraction = 10.
    else:
        if fraction <= 1: nice_fraction = 1.
        elif fraction <= 2: nice_fraction = 2.
        elif fraction <= 5: nice_fraction = 5.
        else: nice_fraction = 10.

    return nice_fraction * 10 ** exponent

def nice_bounds(axis_start, axis_end, num_ticks=10):
    '''
    nice_bounds(axis_start, axis_end, num_ticks=10) -> tuple
    @return: tuple as (nice_axis_start, nice_axis_end, nice_tick_width)
    '''
    axis_width = axis_end - axis_start
    if axis_width == 0:
        nice_tick = 0
    else:
        nice_range = nice_number(axis_width)
        nice_tick = nice_number(nice_range / (num_ticks -1), round_=True)
        axis_start = math.floor(axis_start / nice_tick) * nice_tick
        axis_end = math.ceil(axis_end / nice_tick) * nice_tick

    return axis_start, axis_end, nice_tick

用作：

>>> nice_bounds(26.5, 28.3)
(26.4, 28.4, 0.2)

还添加了一个javascript移植：

function nice_number(value, round_){
    //default value for round_ is false
    round_ = round_ || false;
    // :latex: \log_y z = \frac{\log_x z}{\log_x y}
    var exponent = Math.floor(Math.log(value) / Math.log(10));
    var fraction = value / Math.pow(10, exponent);

    if (round_)
        if (fraction < 1.5)
            nice_fraction = 1.
        else if (fraction < 3.)
            nice_fraction = 2.
        else if (fraction < 7.)
            nice_fraction = 5.
        else
            nice_fraction = 10.
    else
        if (fraction <= 1)
            nice_fraction = 1.
        else if (fraction <= 2)
            nice_fraction = 2.
        else if (fraction <= 5)
            nice_fraction = 5.
        else
            nice_fraction = 10.

    return nice_fraction * Math.pow(10, exponent)
}

function nice_bounds(axis_start, axis_end, num_ticks){
    //default value is 10
    num_ticks = num_ticks || 10;
    var axis_width = axis_end - axis_start;

    if (axis_width == 0){
        axis_start -= .5
        axis_end += .5
        axis_width = axis_end - axis_start
    }

    var nice_range = nice_number(axis_width);
    var nice_tick = nice_number(nice_range / (num_ticks -1), true);
    var axis_start = Math.floor(axis_start / nice_tick) * nice_tick;
    var axis_end = Math.ceil(axis_end / nice_tick) * nice_tick;
    return {
        "min": axis_start,
        "max": axis_end,
        "steps": nice_tick
    }
}

Answer 3

下面是我自动计算滴答的python代码，它需要数据范围和最大滴答数。

例如：

auto_tick([-120, 580], max_tick=10, tf_inside=False)
Out[224]: array([-100.,   -0.,  100.,  200.,  300.,  400.,  500.])
auto_tick([-120, 580], max_tick=20, tf_inside=False)
Out[225]: array([-100.,  -50.,   -0.,   50.,  100.,  150.,  200.,  250.,  300., 350.,  400.,  450.,  500.,  550.])

以下是函数的Python代码

def auto_tick(data_range, max_tick=10, tf_inside=False):
    """
    tool function that automatically calculate optimal ticks based on range and the max number of ticks
    :param data_range:   range of data, e.g. [-0.1, 0.5]
    :param max_tick:     max number of ticks, an interger, default to 10
    :param tf_inside:    True/False if only allow ticks to be inside
    :return:             list of ticks
    """
    data_span = data_range[1] - data_range[0]
    scale = 10.0**np.floor(np.log10(data_span))    # scale of data as the order of 10, e.g. 1, 10, 100, 0.1, 0.01, ...
    list_tick_size_nmlz = [5.0, 2.0, 1.0, 0.5, 0.2, 0.1, 0.05, 0.02, 0.01]   # possible tick sizes for normalized data in range [1, 10]
    tick_size_nmlz = 1.0     # initial tick size for normalized data
    for i in range(len(list_tick_size_nmlz)):                 # every loop reduces tick size thus increases tick number
        num_tick = data_span/scale/list_tick_size_nmlz[i]     # number of ticks for the current tick size
        if num_tick > max_tick:                               # if too many ticks, break loop
            tick_size_nmlz = list_tick_size_nmlz[i-1]
            break
    tick_size = tick_size_nmlz * scale             # tick sizse for the original data
    ticks = np.unique(np.arange(data_range[0]/tick_size, data_range[1]/tick_size).round())*tick_size    # list of ticks

    if tf_inside:     # if only allow ticks within the given range
        ticks = ticks[ (ticks>=data_range[0]) * (ticks<=data_range[1])]

    return ticks

Answer 4

这是TypeScript / JavaScript ES6中@uesp answer的版本：

function niceNumber(value: number, round = false) {

  const exponent = Math.floor(Math.log10(value));
  const fraction = value / Math.pow(10, exponent);

  let niceFraction: number;

  if (round) {
    if (fraction < 1.5) {
      niceFraction = 1.0;
    } else if (fraction < 3.0) {
      niceFraction = 2.0;
    } else if (fraction < 7.0) {
      niceFraction = 5.0;
    } else {
      niceFraction = 10.0;
    }
  } else {
    if (fraction <= 1.0) {
      niceFraction = 1.0;
    } else if (fraction <= 2.0) {
      niceFraction = 2.0;
    } else if (fraction <= 5.0) {
      niceFraction = 5.0;
    } else {
      niceFraction = 10.0;
    }
  }

  return niceFraction * Math.pow(10, exponent);

}

export function calcTicks(minValue: number, maxValue: number, ticksCount: number) {

  const range = niceNumber(maxValue - minValue);

  const tickValue = niceNumber(range / (ticksCount - 1), true);

  const ticks: number[] = [];
  for (let i = 0; i < ticksCount; i++) {
    ticks.push(minValue + tickValue * i);
  }

  return ticks;

}

但是，这里的calcTicks()函数返回的是刻度线数组，而不是开始和结束范围。

智能计算图表刻度位置

4 个答案: