我有一个很长的功能，所以请耐心等待...我基本上试图根据输入的组合找到2或3个多项式曲线下的总面积。
类似this StackO帖子，但涉及多个多项式曲线。
在整个比赛中，驾驶员有各种轮胎组合和限制长度（在轮胎上花费的时间）。
EG。让我们说一场比赛有10圈。[＆＃39; Super soft＆＃39;，＆＃39; Super soft＆＃39;，＆＃39; Super soft＆＃39;，＆＃39; Super soft＆＃39;，＆ ＃39; Soft＆＃39;，＆＃34; Soft＆＃34;，＆＃34; Soft＆＃34;，＆＃34; Soft＆＃34;，＆＃34; Soft＆＃34;，Soft＆＃34;，＆ ＃34;柔软]，意味着在软胎上使用了超级软胎，6圈（第2次）4圈（第1次）。
我有一个df，其中包含每个场景的坐标集（例如，Tire：Super soft，Stint：1）。现在，我想根据这些系数的总和计算总时间。下面的函数会产生一个结果，但结果与我的手动求和不符。
def calc_optimal_(tyre_to_use_list, name, year, tyre_degrad_models):
timings = []
tyre_strategy = []
freqs = []
strategy = []
# Filter tyre degradation models for the race
model = tyre_degrad_models[tyre_degrad_models['name'] == name].reset_index(drop=True)
#model = model[model["coeffs new"] != "Did not run a full stint on this tyre during the race"]
# Generate tyre permutations of Super soft and Soft
combis_all = tyre_combis(tyre_to_use_list, name , year)
# Ensures that each tyre name appears once
if len(tyre_to_use_list) == 2:
combis = list(filter(lambda x: len(set(x)) > 1, combis_all))
elif len(tyre_to_use_list) == 3:
combis = list(filter(lambda x: len(set(x)) > 2, combis_all))
# Delete integer at end of tyre combination if there are multiple names of the same combination
def remove_int(x):
return x.translate(None, digits).rstrip(" ")
combis_new = [map(remove_int, combis[x]) for x in range(len(combis))]
tyre_to_use_list = map(remove_int, tyre_to_use_list)
# THIS IS THE PART WHERE I AM STUCK AT. Function to calculate total race time if tyre combination was used
def calc_time_per_combi(x):
freq = [len(list(group)) for key, group in groupby(combis[x])]
timing = []
summed = 0
f1 = model[(model['tyre'] == combis_new[x][freq[0]]) & (model['stint'] == 0+1)]['coeffs centered'].reset_index(drop=True)[0]
f1 = f1[:freq[0]]
t1 = simps(f1, dx=5)
timing.append(t1)
print timing
if len(freq) > 1:
f2 = model[(model['tyre'] == combis_new[x][freq[1]]) & (model['stint'] == 1+1)]['coeffs centered'].reset_index(drop=True)[0]
if f2.any():
f2 = f2[0:freq[1]]
t2 = simps(f2, dx=5)
timing.append(t2)
print timing
if len(freq) > 2:
f3 = model[(model['tyre'] == combis_new[x][freq[2]]) & (model['stint'] == 2+1)]['coeffs centered'].reset_index(drop=True)[0]
if f3.any():
f3 = f3[0:freq[2]]
t3 = simps(f3, dx=5)
timing.append(t3)
print timing
summed = sum([int(i) for i in timing])
del timing[ 0:len(timing) ]
return summed
# Create dataframe of results of all combinations
# Loop through all the possible tyre combinations and calculate the total race time for each combi
for x in range(len(combis)):
c = calc_time_per_combi(x)
timings.append(c)
tyre_strategy.append(combis[x])
freq = [len(list(group)) for key, group in groupby(combis[x])]
freqs.append(freq)
df = pd.DataFrame({"tyre strategy": combis,
"tyre freq": freqs,
"timings": timings}).sort_values("timings", ascending=True)
return df
当我使用calc_optimal_timeadv1(['Super soft 1', 'Super soft 2'], "Australian Grand Prix", 2016, all_centered, show_table=True)执行该函数时，"print timing"语句输出此信息，表明列表仍然存储旧结果，每次运行该函数时都不会重置为0，尽管我在return语句之前放置了"del timing[ 0:len(timing) ]"语句。
[2499256.5371356499]
[2499256.5371356499, 0.0]
[2352207.5371378958]
[2352207.5371378958, -3366.5472821174626]
[2211012.4066790882]
[2211012.4066790882, -13373.456908395456]
tyre_degrad_models dataframe
{'coeffs new': {0: array([ 0.00000000e+00, 2.25236029e+02, 3.61353708e+02,
4.08353035e+02, 3.66234012e+02, 2.34996638e+02,
1.46409138e+01, -2.94833161e+02, -6.93425587e+02,
-1.18113636e+03, -1.75796549e+03, -2.42391297e+03,
-3.17897880e+03, -4.02316298e+03, -4.95646551e+03,
-5.97888639e+03, -7.09042562e+03, -8.29108320e+03,
-9.58085914e+03, -1.09597534e+04, -1.24277661e+04,
-1.39848970e+04, -1.56311464e+04, -1.73665141e+04,
-1.91910001e+04, -2.11046045e+04, -2.31073272e+04,
-2.51991683e+04, -2.73801278e+04, -2.96502056e+04,
-3.20094017e+04, -3.44577162e+04, -3.69951490e+04,
-3.96217002e+04, -4.23373698e+04, -4.51421577e+04,
-4.80360639e+04, -5.10190885e+04, -5.40912315e+04,
-5.72524928e+04, -6.05028724e+04, -6.38423704e+04,
-6.72709868e+04, -7.07887215e+04, -7.43955745e+04,
-7.80915459e+04, -8.18766357e+04, -8.57508438e+04,
-8.97141702e+04, -9.37666150e+04, -9.79081782e+04,
-1.02138860e+05, -1.06458660e+05, -1.10867578e+05,
-1.15365614e+05, -1.19952769e+05, -1.24629042e+05]),
1: array([ 0.00000000e+00, 9.38647679e+01, 1.67183908e+02,
2.19957421e+02, 2.52185306e+02, 2.63867564e+02,
2.55004193e+02, 2.25595196e+02, 1.75640571e+02,
1.05140318e+02, 1.40944372e+01, -9.74970708e+01,
-2.29634206e+02, -3.82316970e+02, -5.55545360e+02,
-7.49319379e+02, -9.63639025e+02, -1.19850430e+03,
-1.45391520e+03, -1.72987173e+03, -2.02637388e+03,
-2.34342167e+03, -2.68101508e+03, -3.03915412e+03,
-3.41783879e+03, -3.81706908e+03, -4.23684500e+03,
-4.67716655e+03, -5.13803373e+03, -5.61944653e+03,
-6.12140497e+03, -6.64390903e+03, -7.18695871e+03,
-7.75055403e+03, -8.33469497e+03, -8.93938154e+03,
-9.56461374e+03, -1.02103916e+04, -1.08767150e+04,
-1.15635841e+04, -1.22709988e+04, -1.29989591e+04,
-1.37474651e+04, -1.45165167e+04, -1.53061139e+04,
-1.61162568e+04, -1.69469452e+04, -1.77981793e+04,
-1.86699591e+04, -1.95622844e+04, -2.04751554e+04,
-2.14085720e+04, -2.23625343e+04, -2.33370421e+04,
-2.43320956e+04, -2.53476947e+04, -2.63838395e+04])},
'name': {0: 'Australian Grand Prix', 1: 'Australian Grand Prix'},
'stint': {0: 1.0, 1: 2.0},
'tyre': {0: u'Super soft', 1: u'Super soft'},
'year': {0: 2016, 1: 2016}}