我有這個資料：
puf = pd.DataFrame({'id':[1,2,3,4,5,6,7,8],
'val':[850,1889,3289,6083,10349,17860,28180,41236]})
資料似乎遵循指數曲線。讓我們看看劇情：
puf.plot('id','val')
我想擬合指數曲線（$$ y = Ae^{Bx} $$，A乘以e到B * X）并將其添加為Pandas中的一列。首先，我嘗試記錄值：
puf['log_val'] = np.log(puf['val'])
然后使用 Numpy 來擬合方程：
puf['fit'] = np.polyfit(puf['id'],puf['log_val'],1)
但我收到一個錯誤：
ValueError: Length of values (2) does not match length of index (8)
我的預期結果是將擬合值作為 Pandas 中的新列。我附上了我想要的列擬合值的影像（橙色）：
我被困在這段代碼中。我不確定我做錯了什么。如何使用我的擬合值創建一個新列？
uj5u.com熱心網友回復：
Note that you asked for an exponential model yet you have the results for log-linear model.
Check out the work below:
For log-linear, we are fitting E(log(Y))ie log(y) - (log(b[0]) b[1]*x):
from scipy.optimize import least_squares
least_squares(lambda b: np.log(puf['val']) -(np.log(b[0]) b[1] * puf['id']),
[1,1])['x']
array([5.99531305e 02, 5.51106793e-01])
These are the values that excel gives.
On the other hand to fit an exponential curve, the randomness is on Y and not on its logarithm, E(Y)=b[0]*exp(b[1] *x) Hence we have:
least_squares(lambda b: puf['val'] - b[0]*exp(b[1] * puf['id']), [0,1])['x']
array([1.08047304e 03, 4.58116127e-01]) # correct results for exponential fit
Depending on your model choice, the values are alittle different.
Better Model? Since you have same number of parameters, consider the one that gives you lower deviance or better out of sample prediction
Note that the ideal exponential model is E(Y) = A'B'^X which for comparison can be written as log(E(Y)) = A XB while log-linear model will be E(log(Y) = A XB. Note the difference in Expectation.
From the two models we have:
Notice how when we go to higher numbers the log-linear overestimates. While in the lower numbers the exponential overestimates.
Code for image:
from scipy.optimize import least_squares
log_lin = least_squares(lambda b: np.log(puf['val']) -(np.log(b[0]) b[1] * puf['id']),
[1,1])['x']
expo = least_squares(lambda b: puf['val'] - b[0]*exp(b[1] * puf['id']), [0,1])['x']
exp_fun = lambda x: expo[0] * exp(expo[1]*x)
log_lin_fun = lambda x:log_lin[0] * exp(log_lin[1]*x)
plt.plot(puf.id, puf.val, label = 'original')
plt.plot(puf.id, exp_fun(puf.id), label='exponential')
plt.plot(puf.id, log_lin_fun(puf.id), label='log-linear')
plt.legend()
uj5u.com熱心網友回復：
Your getting that error because np.polyfit(puf['id'],puf['log_val'],1) returns two values array([0.55110679, 6.39614819]) which isn't the shape of your dataframe.
This is what you want
y = a* exp (b*x) -> ln(y)=ln(a) bx
f = np.polyfit(df['id'], np.log(df['val']), 1)
where
a = np.exp(f[1]) -> 599.5313046712091
b = f[0] -> 0.5511067934637022
Giving
puf['fit'] = a * np.exp(b * puf['id'])
id val fit
0 1 850 1040.290193
1 2 1889 1805.082864
2 3 3289 3132.130026
3 4 6083 5434.785677
4 5 10349 9430.290286
5 6 17860 16363.179739
6 7 28180 28392.938399
7 8 41236 49266.644002
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