如何让python添加和划分多个变量?

Question

问候 python 程序员。我正在构建一个树莓派项目，旨在在我的花园中使用湿度传感器。我的程序设计为获取三十秒的读数，然后将它们平均为一个数字。它使用四个传感器并运行python3.5传感器都工作得很好，但是我在让python进行数学运算以平均输入时遇到了问题。我认为它正在添加变量，但似乎并没有划分它们。这是代码：

TEST_SOIL()
outpuT = str(output)
print("Moisture:", outpuT)
GPIO.output(buzzer,GPIO.HIGH)
sleep(0.1) # Delay in seconds
GPIO.output(buzzer,GPIO.LOW)
sleep(0.1)
GPIO.output(buzzer,GPIO.HIGH)
sleep(0.1) # Delay in seconds
GPIO.output(buzzer,GPIO.LOW)
print("TESTING SOIL")
o1 = (output + output2 + output3 + output4/4)
sleep(1)
TEST_SOIL()
o2 = (output + output2 + output3 + output4/4)
sleep(1)
TEST_SOIL()
o3 = (output + output2 + output3 + output4/4)
sleep(1)
TEST_SOIL()
o4 = (output + output2 + output3 + output4/4)
sleep(1)
TEST_SOIL()
o5 = (output + output2 + output3 + output4/4)
sleep(1)
TEST_SOIL()
o6 = (output + output2 + output3 + output4/4)
sleep(1)
TEST_SOIL()
o7 = (output + output2 + output3 + output4/4)
sleep(1)
TEST_SOIL()
o8 = (output + output2 + output3 + output4/4)
sleep(1)
TEST_SOIL()
o9 = (output + output2 + output3 + output4/4)
sleep(1)
TEST_SOIL()
o10 = (output + output2 + output3 + output4/4)
sleep(1)
TEST_SOIL()
o11 = (output + output2 + output3 + output4/4)
sleep(1)
TEST_SOIL()
o12 = (output + output2 + output3 + output4/4)
sleep(1)
TEST_SOIL()
o13 = (output + output2 + output3 + output4/4)
sleep(1)
TEST_SOIL()
o14 = (output + output2 + output3 + output4/4)
sleep(1)
TEST_SOIL()
o15 = (output + output2 + output3 + output4/4)
sleep(1)
TEST_SOIL()
o16 = (output + output2 + output3 + output4/4)
sleep(1)
TEST_SOIL()
o17 = (output + output2 + output3 + output4/4)
sleep(1)
TEST_SOIL()
o18 = (output + output2 + output3 + output4/4)
sleep(1)
TEST_SOIL()
o19 = (output + output2 + output3 + output4/4)
sleep(1)
TEST_SOIL()
o20 = (output + output2 + output3 + output4/4)
sleep(1)
TEST_SOIL()
o21 = (output + output2 + output3 + output4/4)
sleep(1)
TEST_SOIL()
o22 = (output + output2 + output3 + output4/4)
sleep(1)
TEST_SOIL()
o23 = (output + output2 + output3 + output4/4)
sleep(1)
TEST_SOIL()
o24 = (output + output2 + output3 + output4/4)
sleep(1)
TEST_SOIL()
o25 = (output + output2 + output3 + output4/4)
sleep(1)
TEST_SOIL()
o26 = (output + output2 + output3 + output4/4)
sleep(1)
TEST_SOIL()
o27 = (output + output2 + output3 + output4/4)
sleep(1)
TEST_SOIL()
o28 = (output + output2 + output3 + output4/4)
sleep(1)
TEST_SOIL()
o29 = (output + output2 + output3 + output4/4)
sleep(1)
TEST_SOIL()
o30 = (output + output2 + output3 + output4/4)
sleep(1)
o = (o1 + o2 + o3 + o4 + o5 + o6 + o7 + o8 + o9 + o10 + o11 + o12 + o13 + o14 + o15 + o16 + o17 + o18 + o19 + o20 + o21 + o22 + o23 + o24 + o25 + o26 + o27 + o28 + o29 + o30)
o = (o/30) 

如果有人想知道"TEST_SOIL(("是什么，那就是我自己在程序前面创建的函数。它的定义如下：

def TEST_SOIL():
output = analogInput(0) # Reading from CH0
output = interp(output, [0, 1023], [100, 0])
output = int(output)
output = (output - 12.5)
output2 = analogInput(1) # Reading from CH0
output2 = interp(output2, [1, 1023], [100, 1])
output2 = int(output2)
output2 = (output2 - 12.5)
output3 = analogInput(2) # Reading from CH0
output3 = interp(output3, [2, 1023], [100, 2])
output3 = int(output3)
output3 = (output3 - 12.5)
output4 = analogInput(3) # Reading from CH0
output4 = interp(output4, [3, 1023], [100, 3])
output4 = int(output4)
output4 = (output4 - 12.5)

我需要注意的是，这段代码还控制一个小蜂鸣器，以消除与"GPIO.output(蜂鸣器，高/低("文本的任何混淆。如果有人能告诉我如何修复此代码，那就太好了。
:)

Answer 1

我冒昧地重构了一下你的代码。

我假设 4 种不同的模拟读数需要这样并且没有碰它们，但除此之外......

• test_soil现在返回读数的 4 元组。
• average_output()需要任何可迭代的数字来取平均值。
• beep()蜂鸣器发出哔哔声。
• main循环调用上述位，将值收集到一个数组中，然后最终平均该数组。

def TEST_SOIL():
output = analogInput(0)  # Reading from CH0
output = interp(output, [0, 1023], [100, 0])
output = int(output)
output = output - 12.5
output2 = analogInput(1)  # Reading from CH0
output2 = interp(output2, [1, 1023], [100, 1])
output2 = int(output2)
output2 = output2 - 12.5
output3 = analogInput(2)  # Reading from CH0
output3 = interp(output3, [2, 1023], [100, 2])
output3 = int(output3)
output3 = output3 - 12.5
output4 = analogInput(3)  # Reading from CH0
output4 = interp(output4, [3, 1023], [100, 3])
output4 = int(output4)
output4 = output4 - 12.5
return (output, output2, output3, output4)

def average_output(o):
return sum(o) / len(o)

def beep():
GPIO.output(buzzer, GPIO.HIGH)
sleep(0.1)  # Delay in seconds
GPIO.output(buzzer, GPIO.LOW)

def main():
ao = average_output(TEST_SOIL())
print("Initial Moisture:", ao)
beep()
beep()
print("TESTING SOIL")
values = []
for x in range(30):
values.append(average_output(TEST_SOIL()))
sleep(1)
final_average = average_output(values)
print("Final Moisture:", final_average )

Answer 2

将(output + output2 + output3 + output4/4)更改为(output + output2 + output3 + output4)/4

您的代码首先将最后一个变量 (ouotput4( 除以 4，然后将它们相加。

(编辑：忘记列表没有平均功能。 对不起。您可以将sum()用于某些列表，然后将其除以此列表的len()(

Answer 3

这也许能够修复您的代码并对其进行一些整理 与其再次编写相同的东西，您可以使用 for 循环使工作更容易

for step in range(30):
TEST_SOIL()
op_avg = (output1 + output2 + output3 + output4) / 4.0
op = op + op_avg
sleep(1)
op = op/30

希望对您有所帮助