I'm trying to get my code to get a string of data from a sensor, and then do something when it reads a specific value.
The following code is how I'm receiving the data now (this is just a test) the function is called earlier in the code at the time where I want it to be called.
def gettemperature(self)
temp = self.board.temp_sensor
print("Temperature is " + str(round(temp)) + " degrees.")
This code works, and returns the temperature value rounded to the terminal, but how could I, instead of printing the value to the terminal, make it so when that string of rounded value is say, 200 degrees, then it prints the temperature value? instead of printing it every 2 seconds (the frequency of the data being received, as per another part of the code)
Using something like
if temp = 200
then print(blahblah)
in short, the above code is what I'm trying to do. If the temp equals a certain value, then something else will happen.
That last code doesn't work. I'm pretty new to coding, so I'm assuming I'm either not going about this the right way, or the syntax of how I'm going about trying to get that value to do something isn't correct (obviously)
Thanks for any help! I'm surprised I got this far, haha.
It would be better if your function gettemperature would return something and then print the result in the condition:
def gettemperature()
temp = board.temp_sensor
return temp
temp = gettemperature()
if temp == 200:
print("Temperature is " + str(round(temp)) + " degrees.")
Before using stackoverflow, I'd recommend learning all this stuff from some basic course, as you'll get to learn the stuff yourself rather then get the answer from someone else.
Try learning conditional statements.
what you want is, to put a conditional statement which triggers if temperature is greater than 200.
If the temp is always a number in string data type, you can use the below code.
def gettemperature(self):
temp = self.board.temp_sensor
print("Temperature is " + str(round(temp)) + " degrees.")
temp=int(temp) #typecasting string datatype to integer
if temp == 200:
print("Temperature is high")
Related
I might be asking a simple question. I have a python program that runs every minute. But I would like a block of code to only run once the condition changes? My code looks like this:
# def shortIndicator():
a = int(indicate_5min.value5)
b = int(indicate_10min.value10)
c = int(indicate_15min.value15)
if a + b + c == 3:
print("Trade posible!")
else:
print("Trade NOT posible!")
# This lets the processor work more than it should.
"""run_once = 0 # This lets the processor work more than it should.
while 1:
if run_once == 0:
shortIndicator()
run_once = 1"""
I've run it without using a function. But then I get an output every minute. I've tried to run it as a function, when I enable the commented code it sort of runs, but also the processing usage is more. If there perhaps a smarter way of doing this?
It's really not clear what you mean, but if you only want to print a notification when the result changes, add another variable to rembember the previous result.
def shortIndicator():
return indicate_5min.value5 and indicate_10min.value10 and indicate_15min.value15
previous = None
while True:
indicator = shortIndicator()
if previous is None or indicator != previous:
if indicator:
print("Trade possible!")
else:
print("Trade NOT possible!")
previous = indicator
# take a break so as not to query too often
time.sleep(60)
Initializing provious to None creates a third state which is only true the first time the while loop executes; by definition, the result cannot be identical to the previous result because there isn't really a previous result the first time.
Perhaps also notice the boolean shorthand inside the function, which is simpler and more idiomatic than converting each value to an int and checking their sum.
I'm guessing the time.sleep is what you were looking for to reduce the load of running this code repeatedly, though that part of the question remains really unclear.
Finally, check the spelling of possible.
If I understand it correctly, you can save previous output to a file, then read it at the beginning of program and print output only if previous output was different.
Current assignment is building a basic text adventure. I'm having trouble with the following code. The current assignment uses only functions, and that is the way the rules of the assignment state it must be done.
def make_selections(response):
repeat = True
while repeat == True:
selection = raw_input('-> ')
for i, v in enumerate(response):
i +=1 # adds 1 to the index to make list indices correlate to a regular 1,2,3 style list
if selection == i:
print v[1]
else:
print "There's an error man, what are you doing?!?!?"
firstResponse = 'You chose option one.'
secondResponse = 'You chose option two.'
thirdResponse = 'You chose option three.'
responses = [(0, firstResponse), (1, secondResponse),( 0, thirdResponse)]
make_selections(responses)
My intention in that code is to make it so if the user selects a 1, it will return firstResponse, if the user selects 2 it will return secondResponse, etc.
I am basically just bug testing the code to make sure it produces the appropriate response, hence the "Error man..." string, but for some reason it just loops through the error message without printing the appropriate response string. Why is this?
I know that this code is enumerating the list of tuples and I can call them properly, as I can change the code to the following and get the expected output:
for i, v in enumerate(response):
i += 1 # adds 1 to the index to make list indices correlate to a regular 1,2,3 style list
print i, v
Also, two quick asides before anyone asks:
I know there is currently no way to get out of this while loop. I'm just making sure each part of my code works before I move on to the next part. Which brings me to the point of the tuples.
When I get the code working, a 0 will produce the response message and loop again, asking the user to make a different selection, whereas a 1 will produce the appropriate response, break out of the loop, and move on to the next 'room' in the story... this way I can have as many 'rooms' for as long of a story as I want, the player does not have to 'die' each time they make an incorrect selection, and each 'room' can have any arbitrary amount of options and possible responses to choose from and I don't need to keep writing separate loops for each room.
There are a few problems here.
First, there's no good reason to iterate through all the numbers just to see if one of them matches selection; you already know that will be true if 1 <= selection <= len(response), and you can then just do response[selection-1] to get the v. (If you know anything about dicts, you might be able to see an even more convenient way to write this whole thing… but if not, don't worry about it.)
But if you really want to do this exhaustive search, you shouldn't print out There is an error man after any mismatch, because then you're always going to print it at least twice. Instead, you want to only print it if all of them failed to match. You can do this by keeping track of a "matched" flag, or by using a break and an else: clause on your for loop, whichever seems simpler, but you have to do something. See break and continue Statements, and else Clauses on Loops in the tutorial for more details.
But the biggest problem is that raw_input returns a string, and there's no way a string is ever going to be equal to a number. For example, try '1' == 1 in your interactive interpreter, and it'll say False. So, what you need to do is convert the user's input into a number so you can compare it. You can do that like this:
try:
selection = int(selection)
except ValueError:
print "That's not a number!"
continue
Seems like this is a job for dictionaries in python. Not sure if your assignment allows this, but here's my code:
def make_selections(response):
selection = raw_input('-> ')
print response.get(selection, err_msg)
resp_dict = {
'1':'You chose option one.',
'2':'You chose option two.',
'3':'You chose option three.'
}
err_msg = 'Sorry, you must pick one of these choices: %s'%sorted(resp_dict.keys())
make_selections(resp_dict)
The problem is that you are comparing a string to an integer. Selection is raw input, so it comes in as a str. Convert it to an int and it will evaluate as you expect.
You can check the type of a variable by using type(var). For example, print type(selection) after you take the input will return type 'str'.
def make_selections(response):
repeat = True
while repeat == True:
selection = raw_input('-> ')
for i, v in enumerate(response):
i +=1 # adds 1 to the index to make list indices correlate to a regular 1,2,3 style list
if int(selection) == i:
print v[1]
else:
print "There's an error man, what are you doing?!?!?"
I spent a good hour or more looking for the answer on here. I have found a few things that help, but do not answer my question specifically. I am using Python 3.3.3. I am a novice so please be gentle.
I am trying to create a program that takes a user input, but then I need to do a check to see what datatype that input is, and then based on that datatype take a certain course of action.
Any string besides those found in this list:
valid_help_string_list = ['\'help\'', '\'HELP\'', 'help', 'HELP']
should result in the printing of:
'please enter a valid entry' or something to that effect.
Any integer (over 0 but under 500) should have float() used on it to make the rows line up.
Any float (over 0.0 but under 500.0) is valid.
For the sake of this project I am assuming nobody using this will weigh under 100 lbs or over 500.
Anything not falling within those categories should also yield the same "please enter a valid response" error message to the user.
I think it's simple enough of a project to take on for a novice. The program is meant to allow you to input your weight and then creates a pictogram based on that weight and saves it all on the next open line of the .txt file I have set up for it. Or if you want to see the legend for the pictogram, you should be able to type help in any variation found in that list.
Any help would be much appreciated.
The user input will be a string by default, so we need to check whether it could become an integer or float. As you want to turn the integers in floats anyway, there's no need to do anything complex:
def validate_input(val, min_v=100, max_v=500):
try:
val = float(val)
except ValueError:
print("Not a valid entry")
else:
if not min_v < val <= max_v:
print("Value should be between {} and {}".format(min_v, max_v))
else:
return val
return False
Now your calling loop can read:
while True:
val = input("...")
if val in valid_help_string_list:
# print help
else:
val = validate_input(val)
if val:
break
# use val
Note that this relies on the return from validate_input being either False or a number larger than 0; Python will interpret a zero return as False and not reach the break, so I recommend keeping min_v >= 0.
I have a function in Python that reads a ds18b20 temp sensor. The sensor gives me a faulty value (-0.062) about 5% of the time that I read it. This is not a problem but I do not want to log the value since it looks ugly in my graphs.
I can't manage to "catch" the value in an if-statement to replace it with "#error". The code below runs nicely but it appears that the if-statement is faulty and does not work - it just runs everything under the else.
I have tried everything, even "catching" all values between 1000 and 1500 (present temperature reading before dividing by 1000) to check if it would work with any temperature, but it does not.
Does anyone have any idea why my if-statement does not work?
def readtwo():
tfile = open("/sys/bus/w1/devices/28-0000040de8fc/w1_slave")
text = tfile.read()
tfile.close()
secondline = text.split("\n")[1]
temperaturedata = secondline.split(" ")[9]
temperature = float(temperaturedata[2:])
temperature = temperature / 1000
if temperature == -0.062:
return("#error")
else:
return(temperature)
Testing base 10 floats for (in)equality is almost always the wrong thing to do, because they almost always cannot be exactly represented in a binary system.
From what I see of your snippet, you should compare against the string, then convert to float if it is not the dreaded -0.062:
def readtwo():
tfile = open("/sys/bus/w1/devices/28-0000040de8fc/w1_slave")
text = tfile.read()
tfile.close()
secondline = text.split("\n")[1]
temperaturedata = secondline.split(" ")[9]
temperature = temperaturedata[2:]
if temperature == '-0062':
return("#error")
else:
temperature = float(temperature) / 1000
return(temperature)
You might also be able to clean up the rest of the code a little:
def readtwo():
with open("/sys/bus/w1/devices/28-0000040de8fc/w1_slave", 'r') as f:
secondline = f.readlines()[1]
temp = secondline.split(' ')[9][2:]
if '-62' in temp:
return '#error'
else:
return float(temp)/1000
Regardless to my comment about the decimal module, floating point arithemitc has it's problems (in python as well). The foremost of which is that due to representation errors, two numbers that are equal on paper, will not be equal when compared by the program.
The way around it, is to look at the relative error between two numbers, rather than simply compare them.
in pseudo:
if abs(num1 - num2)/ abs(num2) < epsilon:
print "They are close enough"
And in your case:
if abs(temparture + 0.062)/0.062 < 10**-3:
return("#error")
Basically, we check that the numbers are "close enough" to be considered the same.
I want to print the result of the equation in my if statement if the input is a digit and print "any thing" if it is a letter.
I tried this code, but it's not working well. What is wrong here?
while 1:
print '\tConvert ciliuse to fehrenhit\n'
temp = input('\nEnter the temp in C \n\t')
f = ((9/5)*temp +32)
if temp.isdigit():
print f
elif temp == "quit" or temp == "q" :
break
elif temp.isalpha() :
print ' hhhhhhh '
You need to go through your code line by line and think about what type you expect each value to be. Python does not automatically convert between, for example, strings and integers, like some languages do, so it's important to keep types in mind.
Let's start with this line:
temp = input('\nEnter the temp in C \n\t')
If you look at the documentation for input(), input() actually calls eval() on what you type in in Python 2.x (which it looks like you're using). That means that it treats what you type in there as code to be evaluated, just the same as if you were typing it in the shell. So if you type 123, it will return an int; if you type 'abc', it will return a str; and if you type abc (and you haven't defined a variable abc), it will give you an error.
If you want to get what the user types in as a string, you should use raw_input() instead.
In the next line:
f = ((9/5)*temp +32)
it looks like you're expecting temp to be a number. But this doesn't make sense. This line gets executed no matter what temp is, and you're expecting both strings containing digits and strings containing letters as input. This line shouldn't go here.
Continuing on:
if temp.isdigit():
isdigit() is a string method, so here you're expecting temp to be a string. This is actually what it should be.
This branch of the if statement is where your equation should go, but for it to work, you will first have to convert temp to an integer, like this:
c = int(temp)
Also, to get your calculation to work out right, you should make the fraction you're multiplying by a floating-point number:
f = ((9/5.0)*c +32)
The rest of your code should be okay if you make the changes above.
A couple of things first - always use raw_input for user input instead of input. input will evaluate code, which is potentially dangerous.
while 1:
print "\tConvert ciliuse to fehrenhit\n"
temp = raw_input("\nEnter the temp in C \n\t")
if temp in ("quit", "q"):
break
try:
f = ((9.0 / 5.0) * float(temp) + 32)
except ValueError:
print "anything"
Instead of using isalpha to check if input is invalid, use a catch clause for ValueError, which is thrown when a non-numerical value is used.
Why isn't it working? Are you getting an error of any kind?
Straight away I can see one problem though. You are doing the calculation before you verify it as a number. Move the calculation to inside the if temp.isdigit().
Take a look at this for some examples:
http://wiki.python.org/moin/Powerful%20Python%20One-Liners
OK, this works. Only problem is when you quit, you get dumped out of the interpreter.
while 1: import sys; temp=raw_input('\nEnter the temp in C \n\t'); temp.isdigit() and sys.stdout.write('%lf' %((9./5)*float(temp)+32)) or temp=='q' and sys.exit(0) or sys.stdout.write(temp)