求曲线下的最大面积| 熊猫，matplotlib

Question

I am struggling a bit in this one - in order to find the necessary battery capacity I need to analyze production/demand over a year's worth of data. To do this, I figured I need to calculate the biggest area under the 0 line. I guess I need to find the start/end points of that area and multiply everything by its respective y-value?

Here is a shortened version of the graph I have: That is the biggest area under the 0 in the image, but in the full dataset it could be any area. I know how to integrate it in the case I find the boundaries of the area in question but I'm struggling to find an efficient way to do that.

My dataframe looks like this:

                     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

I use this snippet to find the length in timesteps of the longest area but I'm missing if there's a way to multiply the points by their y-values (diff). It's technically not correct , however, considering that an area might be long but narrow and another might be shorter and taller, so with an overall bigger area.

def max0(sr):
     return (sr >= 0).cumsum().value_counts().max() - (0 if (sr >= 0).cumsum().value_counts().idxmax() < 0 else 1)

Answer 1

You can find the largest area under the 0-line. I generated my own data

x = np.random.randn(100000)
x = x.cumsum()-x.mean()
plt.plot(x);

Now calculate the start and end points for positive and negative sequences. Every value in a sequence gets an increasing integer to be able to group by sequence.

x1 = np.diff(x < 0).cumsum()

Use pandas groupby to compute all areas and find the largest negative

df = pd.DataFrame({
    'value': x[1:],
    'border': x1
})
dfg = df.groupby('border')
mingr = dfg.apply(lambda x: np.trapz(x.value)).idxmin()
plt.plot(x[1:])
plt.plot(
    dfg.get_group(mingr).value
);
plt.title(
    "position from {} to {}".format(
        dfg.get_group(mingr).index[0],
        dfg.get_group(mingr).index[-1]));

How this works

I create a dataset that is easier to follow along

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--');

I want to know the sequences with consecutive negative or positive values. This can be archived with the filter X < 0.

df = pd.DataFrame({'value': X, 'lt_zero': X < 0})
df[:10]
      value  lt_zero
0  3.125986    False
1  3.885588    False
2  4.580410    False
3  2.998920    False
4  1.913088    False
5 -2.902447     True
6 -3.986654     True
7 -4.373026     True
8 -2.878661     True
9 -1.929964     True

Now I can find the indices where the sign changes, when I diff every consecutive value. I concat one False before the data to not loose the first value.

df['sign_switch'] = np.diff(np.r_[False, X < 0])
df[:10]
      value  lt_zero  sign_switch
0  3.125986    False        False
1  3.885588    False        False
2  4.580410    False        False
3  2.998920    False        False
4  1.913088    False        False
5 -2.902447     True         True
6 -3.986654     True        False
7 -4.373026     True        False
8 -2.878661     True        False
9 -1.929964     True        False

With cumsum() I get for every sequence an increasing integer value. Now I have a grouping variable for every sequence.

df['sign_sequence'] = np.diff(np.r_[False, X < 0]).cumsum()
df[:10]
      value  lt_zero  sign_switch  sign_sequence
0  3.125986    False        False              0
1  3.885588    False        False              0
2  4.580410    False        False              0
3  2.998920    False        False              0
4  1.913088    False        False              0
5 -2.902447     True         True              1
6 -3.986654     True        False              1
7 -4.373026     True        False              1
8 -2.878661     True        False              1
9 -1.929964     True        False              1

For every group I can compute the integral for the values in the group.

sign_groups = df.groupby('sign_sequence')
sign_groups.apply(lambda x: np.trapz(x.value))
sign_sequence
0    13.984455
1   -13.654547
2    14.370044
3   -14.549090

You can access every group later and use the areas. For example to plot the areas.

plt.figure(figsize=(12,5))
plt.plot(X,'o--')
plt.axhline(0, c='gray')
for e,group in enumerate(sign_groups):
    plt.fill_between(group[1].index,0, group[1].value)
    area = np.trapz(group[1].value)
    plt.text((e)*5+1.5, np.sign(area) * 1.25, f'{area:.2f}', fontsize=12)