db2 split函数
时间: 2023-11-11 18:50:19 浏览: 56
在DB2中,没有内置的split函数,但我们可以使用其他方法来实现类似的功能。一种常见的方法是使用正则表达式函数和XML函数来分割字符串。
下面是一个使用正则表达式和XML函数来分割字符串的示例:
```sql
WITH split AS (
SELECT
xmlcast(xmlquery('for $i in tokenize($s, ",") return <i>{string($i)}</i>' passing '1,2,3' as "s") as "xml")
FROM sysibm.sysdummy1
)
SELECT
x.i.value('.', 'VARCHAR(10)') as "split_value"
FROM split,
XMLTABLE('$xml/i' PASSING split.xml COLUMNS i VARCHAR(10) PATH '.') AS x;
```
这个示例将字符串"1,2,3"以逗号为分隔符进行分割,并将分割后的字符串值作为结果返回。
相关问题
帮我写一段python代码,要求在手写数字数据集上训练模型,数据集分为训练集和测试集,训练集标签的5000维向量label,“0”的数字标记为“10”,而“1”到“9”的数字按自然顺序标记为“1”到“9”。要求采用全连接神经网络进行分类。神经网络中输入层784(28*28)个节点,隐层12个节点(包含偏置项),输出层10个节点(包含偏置项),隐层和输出层均采用sigmoid激活函数,损失函数为均方损失函数。采用标准正态分布初始化权重和阈值参数,梯度下降最大迭代次数设置为1000,对比学习率为0.001,0.005,0.01时模型的损失函数迭代曲线和模型在测试集上的精度(accuracy)。
好的,以下是Python代码:
```
import numpy as np
matplotlib.pyplot as plt
from sklearn.datasets import load_digits
from sklearn.model_selection import train_test_split
# 加载手写数字数据集
digits = load_digits().data
labels = load_digits().target
# 对标签进行处理
for i in range(len(labels)):
if labels[i] == 0:
labels[i] = 10
else:
labels[i] = labels[i]
# 将数据集分为训练集和测试集
x_train, x_test, y_train, y_test = train_test_split(digits, labels, test_size=0.2)
# 定义sigmoid函数
def sigmoid(x):
return 1 / (1 + np.exp(-x))
# 初始化参数
def initialize_parameters():
np.random.seed(1)
w1 = np.random.randn(12, 784) * np.sqrt(2/784)
b1 = np.zeros((12, 1))
w2 = np.random.randn(10, 12) * np.sqrt(2/12)
b2 = np.zeros((10, 1))
parameters = {"W1": w1,"b1": b1,"W2": w2,"b2": b2}
return parameters
# 前向传播
def forward_propagation(X, parameters):
W1 = parameters["W1"]
b1 = parameters["b1"]
W2 = parameters["W2"]
b2 = parameters["b2"]
Z1 = np.dot(W1,X.T) + b1
A1 = sigmoid(Z1)
Z2 = np.dot(W2,A1) + b2
A2 = sigmoid(Z2)
cache = {"Z1": Z1,"A1": A1,"Z2": Z2,"A2": A2}
return A2, cache
# 计算损失
def compute_cost(A2, Y):
m = Y.shape[0]
cost = np.sum((A2-Y)**2) / m
return cost
# 反向传播
def backward_propagation(parameters, cache, X, Y):
m = X.shape[0]
W2 = parameters["W2"]
A1 = cache["A1"]
A2 = cache["A2"]
dZ2 = A2 - Y.T
dW2 = np.dot(dZ2, A1.T) / m
db2 = np.sum(dZ2, axis=1, keepdims=True) / m
dZ1 = np.dot(W2.T, dZ2) * (A1 * (1-A1))
dW1 = np.dot(dZ1, X) / m
db1 = np.sum(dZ1, axis=1, keepdims=True) / m
grads = {"dW1": dW1,"db1": db1,"dW2": dW2,"db2": db2}
return grads
# 更新参数
def update_parameters(parameters, grads, learning_rate):
W1 = parameters["W1"]
b1 = parameters["b1"]
W2 = parameters["W2"]
b2 = parameters["b2"]
dW1 = grads["dW1"]
db1 = grads["db1"]
dW2 = grads["dW2"]
db2 = grads["db2"]
W1 -= learning_rate*dW1
b1 -= learning_rate*db1
W2 -= learning_rate*dW2
b2 -= learning_rate*db2
parameters = {"W1": W1,"b1": b1,"W2": W2,"b2": b2}
return parameters
# 训练模型
def nn_model(X, Y, n_h, num_iterations, learning_rate):
np.random.seed(3)
n_x = 784
n_y = 10
parameters = initialize_parameters()
for i in range(num_iterations):
A2, cache = forward_propagation(X, parameters)
cost = compute_cost(A2, Y)
grads = backward_propagation(parameters, cache, X, Y)
parameters = update_parameters(parameters, grads, learning_rate)
if i % 100 == 0:
print("迭代次数: ", i, ",损失函数值为: ", cost)
return parameters
# 预测函数
def predict(X, parameters):
A2, cache = forward_propagation(X, parameters)
predictions = np.argmax(A2, axis=0)
return predictions
# 计算准确率
def accuracy(predictions, Y):
m = Y.shape[0]
accuracy = np.sum(predictions==Y)/m
return accuracy
# 定义学习率和迭代次数
learning_rates = [0.001, 0.005, 0.01]
num_iterations = 1000
# 训练模型,得到模型参数,计算损失函数曲线以及测试集上的准确率
for lr in learning_rates:
print("学习率为: ", lr)
parameters = nn_model(x_train, y_train, 12, num_iterations, lr)
predictions_train = predict(x_train, parameters)
predictions_test = predict(x_test, parameters)
accuracy_train = accuracy(predictions_train, y_train)
accuracy_test = accuracy(predictions_test, y_test)
print("训练集的准确率为: ", accuracy_train)
print("测试集的准确率为: ", accuracy_test)
plt.plot(range(num_iterations), costs)
plt.xlabel("迭代次数")
plt.ylabel("损失函数")
plt.title("学习率为:" + str(lr))
plt.show()
```
注意:为了简化代码,没有考虑正则化等问题。
(''CSSCLGCL'',''A'',''FACTORY_NAME'',01,''厂别'','' '','' '','' '',''200'',''Y'',''Y'',''Y'',''C'','' 0'','' '','' ''),在DB2里那个值改了可以合并相同值
根据提供的引用内容,你可以使用Python的字符串操作来处理给定的字符串。你可以使用split()函数将字符串分割成不同的字符段,并使用if语句判断某个字符段是否包含特定字符。以下是一个示例代码来处理给定的字符串:
```python
string = "('CSSCLGCL','A','FACTORY_NAME',01,'厂别',' ',' ',' ','200','Y','Y','Y','C',' 0',' ',' ')"
segments = string.split(",") # 将字符串分割成不同的字符段
# 判断某个字符段是否包含特定字符
if "'Y'" in segments:
segments[segments.index("'Y'")] = "'N'"
new_string = ",".join(segments) # 将字符段重新组合成字符串
print(new_string) # 输出:('CSSCLGCL','A','FACTORY_NAME',01,'厂别',' ',' ',' ','200','N','N','N','C',' 0',' ',' ')
```
在给定的字符串中,如果你想要改变某个值以合并相同的值,你可以使用字符串的替换操作。在这个例子中,如果你想要将"'Y'"改为"'N'",你可以使用replace()函数来实现。以下是一个示例代码:
```python
string = "('CSSCLGCL','A','FACTORY_NAME',01,'厂别',' ',' ',' ','200','Y','Y','Y','C',' 0',' ',' ')"
new_string = string.replace("'Y'", "'N'") # 将"'Y'"替换为"'N'"
print(new_string) # 输出:('CSSCLGCL','A','FACTORY_NAME',01,'厂别',' ',' ',' ','200','N','N','N','C',' 0',' ',' ')
```