求曲线下最大面积 | 熊猫、matplotlib

Question

我在这方面有点挣扎 - 为了找到必要的电池容量，我需要分析一年的生产/需求数据。为此，我认为需要计算 0 线下方的最大面积。我想我需要找到该区域的起点/终点并将所有内容乘以其各自的 y 值？

这是我的图表的简化版本：

这是图像中 0 下方的最大区域，但在完整数据集中它可以是任何区域。我知道如何在找到相关区域边界的情况下将其集成，但我正在努力寻找一种有效的方法来做到这一点。

我的数据框如下所示：

                     demand  Production    diff

Time

2019-01-01 00:15:01   17.25      32.907  15.657

2019-01-01 00:30:01   17.80      32.954  15.154

...                     ...         ...     ...

2019-01-16 22:15:02   17.34      27.704  10.364

2019-01-16 22:30:01   18.67      35.494  16.824

我使用此代码片段来查找最长区域的时间步长，但我缺少是否有办法将点乘以它们的 y 值（差异）。然而，从技术上讲，这是不正确的，考虑到一个区域可能长而窄，而另一个区域可能又短又高，因此整体区域更大。

def max0(sr):

     return (sr >= 0).cumsum().value_counts().max() - (0 if (sr >= 0).cumsum().value_counts().idxmax() < 0 else 1)

Answer 1

您可以找到 0 线下方的最大面积。我生成了自己的数据x&nbsp;=&nbsp;np.random.randn(100000) x&nbsp;=&nbsp;x.cumsum()-x.mean() plt.plot(x);现在计算正序列和负序列的起点和终点。序列中的每个值都会获得一个递增的整数，以便能够按序列进行分组。x1 = np.diff(x < 0).cumsum()使用 pandas groupby 计算所有区域并找到最大的负数df = pd.DataFrame({&nbsp; &nbsp; 'value': x[1:],&nbsp; &nbsp; 'border': x1})dfg = df.groupby('border')mingr = dfg.apply(lambda x: np.trapz(x.value)).idxmin()plt.plot(x[1:])plt.plot(&nbsp; &nbsp; dfg.get_group(mingr).value);plt.title(&nbsp; &nbsp; "position from {} to {}".format(&nbsp; &nbsp; &nbsp; &nbsp; dfg.get_group(mingr).index[0],&nbsp; &nbsp; &nbsp; &nbsp; dfg.get_group(mingr).index[-1]));这是如何运作的我创建了一个更容易遵循的数据集x = np.array([3,4,4.5,3,2])X = np.r_[x,-x,x,-x]+np.random.normal(0,.2,20)plt.figure(figsize=(12,5))plt.axhline(0, color='gray')plt.plot(X, 'o--');我想知道具有连续负值或正值的序列。这可以使用过滤器 X < 0 进行存档。df = pd.DataFrame({'value': X, 'lt_zero': X < 0})df[:10]&nbsp; &nbsp; &nbsp; value&nbsp; lt_zero0&nbsp; 3.125986&nbsp; &nbsp; False1&nbsp; 3.885588&nbsp; &nbsp; False2&nbsp; 4.580410&nbsp; &nbsp; False3&nbsp; 2.998920&nbsp; &nbsp; False4&nbsp; 1.913088&nbsp; &nbsp; False5 -2.902447&nbsp; &nbsp; &nbsp;True6 -3.986654&nbsp; &nbsp; &nbsp;True7 -4.373026&nbsp; &nbsp; &nbsp;True8 -2.878661&nbsp; &nbsp; &nbsp;True9 -1.929964&nbsp; &nbsp; &nbsp;True现在，当我比较每个连续值时，我可以找到符号发生变化的索引。我在数据之前连接一个 False，以免丢失第一个值。df['sign_switch'] = np.diff(np.r_[False, X < 0])df[:10]&nbsp; &nbsp; &nbsp; value&nbsp; lt_zero&nbsp; sign_switch0&nbsp; 3.125986&nbsp; &nbsp; False&nbsp; &nbsp; &nbsp; &nbsp; False1&nbsp; 3.885588&nbsp; &nbsp; False&nbsp; &nbsp; &nbsp; &nbsp; False2&nbsp; 4.580410&nbsp; &nbsp; False&nbsp; &nbsp; &nbsp; &nbsp; False3&nbsp; 2.998920&nbsp; &nbsp; False&nbsp; &nbsp; &nbsp; &nbsp; False4&nbsp; 1.913088&nbsp; &nbsp; False&nbsp; &nbsp; &nbsp; &nbsp; False5 -2.902447&nbsp; &nbsp; &nbsp;True&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;True6 -3.986654&nbsp; &nbsp; &nbsp;True&nbsp; &nbsp; &nbsp; &nbsp; False7 -4.373026&nbsp; &nbsp; &nbsp;True&nbsp; &nbsp; &nbsp; &nbsp; False8 -2.878661&nbsp; &nbsp; &nbsp;True&nbsp; &nbsp; &nbsp; &nbsp; False9 -1.929964&nbsp; &nbsp; &nbsp;True&nbsp; &nbsp; &nbsp; &nbsp; False我为cumsum()每个序列得到一个递增的整数值。现在我为每个序列都有一个分组变量。df['sign_sequence'] = np.diff(np.r_[False, X < 0]).cumsum()df[:10]&nbsp; &nbsp; &nbsp; value&nbsp; lt_zero&nbsp; sign_switch&nbsp; sign_sequence0&nbsp; 3.125986&nbsp; &nbsp; False&nbsp; &nbsp; &nbsp; &nbsp; False&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 01&nbsp; 3.885588&nbsp; &nbsp; False&nbsp; &nbsp; &nbsp; &nbsp; False&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 02&nbsp; 4.580410&nbsp; &nbsp; False&nbsp; &nbsp; &nbsp; &nbsp; False&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 03&nbsp; 2.998920&nbsp; &nbsp; False&nbsp; &nbsp; &nbsp; &nbsp; False&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 04&nbsp; 1.913088&nbsp; &nbsp; False&nbsp; &nbsp; &nbsp; &nbsp; False&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 05 -2.902447&nbsp; &nbsp; &nbsp;True&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;True&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 16 -3.986654&nbsp; &nbsp; &nbsp;True&nbsp; &nbsp; &nbsp; &nbsp; False&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 17 -4.373026&nbsp; &nbsp; &nbsp;True&nbsp; &nbsp; &nbsp; &nbsp; False&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 18 -2.878661&nbsp; &nbsp; &nbsp;True&nbsp; &nbsp; &nbsp; &nbsp; False&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 19 -1.929964&nbsp; &nbsp; &nbsp;True&nbsp; &nbsp; &nbsp; &nbsp; False&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 1对于每个组，我可以计算组中值的积分。sign_groups = df.groupby('sign_sequence')sign_groups.apply(lambda x: np.trapz(x.value))sign_sequence0&nbsp; &nbsp; 13.9844551&nbsp; &nbsp;-13.6545472&nbsp; &nbsp; 14.3700443&nbsp; &nbsp;-14.549090您可以稍后访问每个组并使用这些区域。例如绘制区域。plt.figure(figsize=(12,5))plt.plot(X,'o--')plt.axhline(0, c='gray')for e,group in enumerate(sign_groups):&nbsp; &nbsp; plt.fill_between(group[1].index,0, group[1].value)&nbsp; &nbsp; area = np.trapz(group[1].value)&nbsp; &nbsp; plt.text((e)*5+1.5, np.sign(area) * 1.25, f'{area:.2f}', fontsize=12)