我之前就同一个问题提出了一个问题，但由于我的方法发生了变化，我现在有不同的问题。
我当前的代码:
from sklearn import preprocessing
from openpyxl import load_workbook
import numpy as np
from numpy import exp, array, random, dot
from sklearn.model_selection import train_test_split
from sklearn.neural_network import MLPRegressor
from sklearn.preprocessing import StandardScaler
from sklearn.metrics import classification_report,confusion_matrix
import matplotlib.pyplot as plt
from sklearn.linear_model import LinearRegression
#Set sizes
rowSize = 200
numColumns = 4
# read from excel file
wb = load_workbook('python_excel_read.xlsx')
sheet_1 = wb["Sheet1"]
date = np.zeros(rowSize)
day = np.zeros(rowSize)
rain = np.zeros(rowSize)
temp = np.zeros(rowSize)
out = np.zeros(rowSize)
for i in range(0, rowSize):
date[i] = sheet_1.cell(row=i + 1, column=1).value
day[i] = sheet_1.cell(row=i + 1, column=2).value
rain[i] = sheet_1.cell(row=i + 1, column=3).value
temp[i] = sheet_1.cell(row=i + 1, column=4).value
out[i] = sheet_1.cell(row=i + 1, column=5).value
train = np.zeros(shape=(rowSize,numColumns))
t_o = np.zeros(shape=(rowSize,1))
for i in range(0, rowSize):
train[i] = [date[i], day[i], rain[i], temp[i]]
t_o[i] = [out[i]]
X = train
# Output
y = t_o
X_train, X_test, y_train, y_test = train_test_split(X, y)
####Neural Net
nn = MLPRegressor(
hidden_layer_sizes=(3,), activation='relu', solver='adam', alpha=0.001, batch_size='auto',
learning_rate='constant', learning_rate_init=0.01, power_t=0.5, max_iter=10000, shuffle=True,
random_state=9, tol=0.0001, verbose=False, warm_start=False, momentum=0.9, nesterovs_momentum=True,
early_stopping=False, validation_fraction=0.1, beta_1=0.9, beta_2=0.999, epsilon=1e-08)
nn.fit(X_train, y_train.ravel())
y_pred = nn.predict(X_test)
###Linear Regression
# lm = LinearRegression()
# lm.fit(X_train,y_train)
# y_pred = lm.predict(X_test)
fig = plt.figure()
ax1 = fig.add_subplot(111)
ax1.scatter(X_test[:,0], y_pred, s=1, c='b', marker="s", label='real')
ax1.scatter(X_test[:,0], y_test, s=10, c='r', marker="o", label='NN Prediction')
plt.show()
#Calc MSE
mse = np.square(y_test-y_pred).mean()
print(mse)
结果显示测试数据的预测非常糟糕。因为我对此很陌生，所以我不确定这是我的数据、模型还是我的编码。根据该图，我认为该模型对于数据来说是错误的(该模型似乎预测了接近线性或平方的数据，而实际数据似乎更加分散)
以下是一些数据点: 格式为一年中的某一天(2 是 1 月 2 日)、工作日(1)/周末(0)、下雨(1)/无雨(0)、F 温度、出勤(这是输出)
2 0 0 51 1366
4 0 0 62 538
5 1 0 71 317
6 1 0 76 174
7 1 0 78 176
8 1 0 68 220
12 1 1 64 256
13 1 1 60 379
14 1 0 64 316
18 0 0 72 758
19 1 0 72 1038
20 1 0 72 405
21 1 0 71 326
24 0 0 74 867
26 1 1 68 521
27 1 0 71 381
28 1 0 72 343
29 1 1 68 266
30 0 1 57 479
31 0 1 57 717
33 1 0 70 542
34 1 0 73 220
35 1 0 74 360
36 1 0 79 444
42 1 0 78 534
45 0 0 80 1572
52 0 0 76 1236
55 1 1 64 689
56 1 0 69 726
59 0 0 67 1188
60 0 0 74 1140
61 1 1 63 979
62 1 1 62 657
63 1 0 67 687
64 1 0 72 615
67 0 0 80 1074
68 1 0 81 1261
71 1 0 83 1332
73 0 0 85 1259
74 0 0 86 1142
76 1 0 88 1207
77 1 1 78 1438
82 1 0 85 1251
83 1 0 83 1019
85 1 0 86 1178
86 0 0 92 1306
87 0 0 92 1273
89 1 0 93 1101
90 1 0 92 1274
93 0 0 83 1548
94 0 0 86 1318
96 1 0 83 1395
97 1 0 81 1338
98 1 0 75 1240
100 0 0 84 1335
102 0 0 83 931
103 1 0 87 746
104 1 0 91 746
105 1 0 81 600
106 1 0 72 852
108 0 1 87 1204
109 0 0 89 1191
110 1 0 90 769
111 1 0 88 642
112 1 0 86 743
114 0 1 75 1085
115 0 1 78 1109
117 1 0 84 871
120 1 0 96 599
123 0 0 93 651
129 0 0 74 1325
133 1 0 88 637
134 1 0 84 470
135 0 1 73 980
136 0 0 72 1096
137 0 0 83 792
138 1 0 87 565
139 1 0 84 501
141 1 0 88 615
142 0 0 79 722
143 0 0 80 1363
144 0 0 82 1506
146 1 0 93 626
147 1 0 94 415
148 1 0 95 596
149 0 0 100 532
150 0 0 102 784
154 1 0 99 514
155 1 0 94 495
156 0 1 87 689
157 0 1 94 931
158 0 0 97 618
161 1 0 92 451
162 1 0 97 574
164 0 0 102 898
165 0 0 104 746
166 1 0 109 587
167 1 0 109 465
174 1 0 108 514
175 1 0 109 572
179 0 0 107 811
181 1 0 104 423
182 1 0 103 526
184 0 1 97 849
185 0 0 103 852
189 1 0 106 728
191 0 0 101 577
194 1 0 105 511
198 0 1 101 616
199 0 1 97 1056
200 0 0 94 740
202 1 0 103 498
205 0 0 101 610
206 0 0 106 944
207 0 0 105 769
208 1 0 103 551
209 1 0 103 624
210 1 0 97 513
212 0 1 107 561
213 0 0 100 905
214 0 0 105 767
215 1 0 107 510
216 1 0 108 406
217 1 0 109 439
218 1 0 103 427
219 0 1 104 460
224 1 0 105 213
227 0 0 112 834
228 0 0 109 615
229 1 0 105 216
230 1 0 104 213
231 1 0 104 256
232 1 0 104 282
235 0 0 104 569
238 1 0 103 165
239 1 1 105 176
241 0 1 108 727
242 0 1 105 652
243 1 1 103 231
244 1 0 96 117
245 1 1 98 168
246 1 1 97 113
247 0 0 95 227
248 0 0 92 1050
249 0 0 101 1274
250 1 1 95 1148
254 0 0 99 180
255 0 0 104 557
258 1 0 94 228
260 1 0 95 133
263 0 0 100 511
264 1 1 89 249
265 1 1 90 245
267 1 0 101 390
272 1 0 100 223
273 1 0 103 194
274 1 0 103 150
275 0 0 95 224
276 0 0 92 705
277 0 1 92 504
279 1 1 77 331
281 1 0 89 268
284 0 0 95 566
285 1 0 94 579
286 1 0 95 420
288 1 0 93 392
289 0 1 94 525
290 0 1 86 670
291 0 1 89 488
294 1 1 74 295
296 0 0 81 314
299 1 0 88 211
301 1 0 84 246
303 0 1 76 433
304 0 0 80 216
307 1 1 80 275
308 1 1 66 319
312 0 0 80 413
313 1 0 78 278
316 1 0 74 305
320 1 1 57 323
324 0 0 76 220
326 0 0 77 461
327 1 0 78 510
331 0 0 60 1701
334 1 0 58 237
335 1 0 62 355
336 1 0 68 266
338 0 0 70 246
342 1 0 72 109
343 1 0 70 103
347 0 0 58 486
349 1 0 52 144
350 1 0 53 209
351 1 0 55 289
354 0 0 62 707
355 1 0 59 903
359 0 0 58 481
360 0 0 53 1342
364 1 0 57 1624
我总共有超过一千个数据点，但我没有将它们全部用于训练/测试。一种想法是我需要更多，另一种想法是我需要更多因素，因为温度/降雨/星期几对出勤率的影响不够。
剧情如下:
我该怎么做才能使我的模型更准确并提供更好的预测？
谢谢
编辑:我添加了更多数据点和另一个因素。我似乎无法上传 Excel 文件，因此我将数据放在这里，并更好地解释其格式
编辑: 这是最新的代码:
from sklearn import preprocessing
from openpyxl import load_workbook
import numpy as np
from numpy import exp, array, random, dot
from sklearn.model_selection import train_test_split
from sklearn.neural_network import MLPRegressor
from sklearn.preprocessing import StandardScaler
from sklearn.metrics import classification_report,confusion_matrix
import matplotlib.pyplot as plt
from sklearn.linear_model import LinearRegression
from sklearn.model_selection import KFold
from sklearn.model_selection import cross_val_predict
from sklearn import svm
from sklearn.model_selection import cross_val_score
from sklearn.metrics import confusion_matrix
from sklearn.preprocessing import StandardScaler
from sklearn.pipeline import Pipeline
from sklearn.model_selection import LeaveOneOut
#Set sizes
rowSize = 500
numColumns = 254
# read from excel file
wb = load_workbook('python_excel_read.xlsx')
sheet_1 = wb["Sheet1"]
input = np.zeros(shape=(rowSize,numColumns))
out = np.zeros(rowSize)
for i in range(0, rowSize):
for j in range(0,numColumns):
input[i,j] = sheet_1.cell(row=i + 1, column=j+1).value
out[i] = sheet_1.cell(row=i + 1, column=numColumns+1).value
output = np.zeros(shape=(rowSize,1))
for i in range(0, rowSize):
output[i] = [out[i]]
X = input
# Output
y = output
print(X)
print(y)
y[y < 500] = 0
y[np.logical_and(y >= 500, y <= 1000)] = 1
y[np.logical_and(y > 1000, y <= 1200)] = 2
y[y > 1200] = 3
# Use cross-validation
#kf = KFold(n_splits = 10, random_state=0)
loo = LeaveOneOut()
# Try different models
clf = svm.SVC()
scaler = StandardScaler()
pipe = Pipeline([('scaler', scaler), ('svc', clf)])
accuracy = cross_val_score(pipe, X, y.ravel(), cv = loo, scoring = "accuracy")
print(accuracy.mean())
#y_pred = cross_val_predict(clf, X, y.ravel(), cv = kf)
#cm = confusion_matrix(y, y_pred)
这是最新的数据，其中包含我可以添加的尽可能多的功能。请注意，这是完整数据中的随机样本:
当前输出: 0.6230954290296712
我的最终目标是达到 90% 或更高的准确率...我不相信我能找到更多的功能，但如果有帮助，我会继续收集尽可能多的功能
最佳答案
你的问题很笼统，但我有一些建议。您可以使用交叉验证并尝试不同的模型。就我个人而言，我会尝试 SVR、RandomForests，最后选择我会使用 MLPR。
我修改了你的代码以显示一个简单的示例:
import numpy as np
from sklearn.preprocessing import StandardScaler
from sklearn.metrics import classification_report,confusion_matrix
import matplotlib.pyplot as plt
from sklearn.linear_model import LinearRegression
from sklearn.model_selection import KFold
from sklearn.model_selection import cross_val_predict
from sklearn import svm
from sklearn.model_selection import cross_val_score
from sklearn.metrics import confusion_matrix
from sklearn.preprocessing import StandardScaler
from sklearn.pipeline import Pipeline
from sklearn.model_selection import LeaveOneOut
import pandas as pd
from sklearn.decomposition import PCA
# read the data
df = pd.read_excel('python_excel_read.xlsx', header = None)
rows, cols = df.shape
X = df.iloc[: , 0:(cols - 1)]
y = df.iloc[: , cols - 1 ]
print(X.shape)
print(y.shape)
y[y < 500] = 0
y[np.logical_and(y >= 500, y <= 1000)] = 1
y[np.logical_and(y > 1000, y <= 1200)] = 2
y[y > 1200] = 3
print(np.unique(y))
# We can apply PCA to reduce the dimensions of the data
# pca = PCA(n_components=2)
# pca.fit(X)
# X = pca.fit_transform(X)
# Use cross-validation
#kf = KFold(n_splits = 10, random_state=0)
loo = LeaveOneOut()
# Try different models
clf = svm.SVC(kernel = 'linear')
scaler = StandardScaler()
pipe = Pipeline([('scaler', scaler), ('svc', clf)])
accuracy = cross_val_score(pipe, X, y.ravel(), cv = loo, scoring = "accuracy")
print(accuracy.mean())
#y_pred = cross_val_predict(clf, X, y.ravel(), cv = kf)
#cm = confusion_matrix(y, y_pred)
关于python - Sklearn NN 回归 出勤预测，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/50847386/