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def getidxmax(window_2, Filtered):
try: bar = window_2[Filtered].idxmax()
return bar
except ValueError:
return np.nan
Am calling the above function like this:
bar = getidxmax(window_2, Filtered)
But am getting error:
File "<algorithm>", line 22
return bar
^
IndentationError: unexpected indent
line 22 is where the try: begins.
Your syntax for try except isn't perfectly correct.
def getidxmax(window_2, Filtered):
try:
return (window_2[Filtered]).idxmax()
except ValueError:
return np.nan
You don't need to create a new variable bar, you can just try and return the result of window_2[Filtered].idmax() and if that fails, np.nan
def getidxmax(window_2, Filtered):
try:
bar = window_2[Filtered].idxmax()
return bar
except ValueError:
return np.nan
Related
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from replit import db
import os
def clear():
del db["hits"]
del db["ab"]
def save():
db["hits"] = hits
db["ab"] = ab
try:
hits = db["hits"]
ab = db["ab"]
avg = hits/ab
print(f"You're currently batting {avg}")
except:
hits = 0
ab = 0
input = input("press ENTER to continue.")
os.system("clear")
if input == clear:
clear()
else:
#the line of code below is the one in question
today_ab = input("How many at bats did you have today?")
ab += today_ab
The last line is where it is wrong.
Python doesn't have separate namespaces for variables and functions. This line
input = input("press ENTER to continue.")
replaces the built-in function input with the result of the call. When you try to call input later, you get the TypeError.
Use a different variable name:
x = input("press ENTER to continue.")
if x == "clear":
clear()
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When running this program I get this error message: NameError: name 'wordList' is not defined. Can't figure out why. Thanks.
import ahocorasick
def build_actree(wordlist):
actree = ahocorasick.Automaton()
for index, word in enumerate(wordlist):
actree.add_word(word, (index, word))
actree.make_automaton()
return actree
if __name__ == '__main__':
actree = build_actree(wordlist=wordlist)
sent_cp = sent
for i in actree.iter(sent):
sent_cp = sent_cp.replace(i[1][1], "**")
sent = '我草你妈'
wordlist =['我草']
print("屏蔽词：",i[1][1])
print("屏蔽结果：",sent_cp)
`
The first thing you do in your code is :
actree = build_actree(wordlist=wordlist)
But there is no place where wordlist is defined. May be some code is missing.
This is the solution I finally came up with:
import ahocorasick
def build_actree(wordlist):
actree = ahocorasick.Automaton()
for index, word in enumerate(wordlist):
actree.add_word(word, (index, word))
actree.make_automaton()
return actree
if __name__ == '__main__':
# correction : add wordlist initialisation
sent = '我草你妈'
wordlist =['我草']
actree = build_actree(wordlist=wordlist)
sent_cp = sent
for i in actree.iter(sent):
sent_cp = sent_cp.replace(i[1][1], "**")
print("屏蔽词：", i[1][1])
print("屏蔽结果：", sent_cp)
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I'm trying to learn how to code really simple neural networks from scratch. However I can't seen to get the training right. Whenever I run my while loop it gets stuck in an infinite loop.
Here is my code:
#inputs and targets
x1=[1,0,1]
x2=[0,1,0]
x3=[0,0,1]
inputs=[x1,x2,x3]
targets=[1,0,0,1]
#parameters
np.random.seed(1)
w1 = np.random.random_sample(size = 3)
w1=w1.tolist()
alpha=0.1
itera=0
#activation function
def purelin(z):
return z
#network
def red(i,w1):
n=[]
for inp in i:
a=inp*w1[inputs.index(i)]
n.append(a)
n=sum(n) #sumar inputs*weights (caso sin bias)
a=purelin(n)
return a
#forward propagation
def forw(inputs):
outputs=[]
for i in inputs:
x=red(i,w1)
outputs.append(x)
return outputs
#error
def error(targets, outputs):
e=[]
zip_object = zip(targets, outputs)
for i, j in zip_object:
e.append(i-j)
return e
#backpropagation
def back(e):
w1_=[]
zip_object2 = zip(w1, e)
for i, j in zip_object2:
w1_.append(i-alpha*j)
return w1_
#training
while itera<1000:
outputs=forw(inputs)
e=error(targets,outputs)
w1=back(e)
itera=+1
I know I still have a lot of work to do but I want to solve this problem so I can tune my code and get results. Thank you!!
This is just a minor typing error. At the end of the loop, you aren't incrementing itera by one, but are setting it to +1 .
Just change the last line to itera += 1and it should work fine.
Note: Sometimes it may be helpful to print out variables, even if I would always recommend you to debug your code properly.
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I've actually already found a work around for my problem, but I feel as though there is still a better way. I can't seem to grasp why the program thinks I'm dealing with a method vs a float.
If I try to run the program, it says that I'm trying to add together two methods. However, when I run the line:
print(type(my.coins.add_coins()))
It tells me that it has returned a float.
Here's the code:
class Currency:
def __init__(self, pennies, nickles, dimes, quarters):
self.pennies = pennies
self.nickles = nickles
self.dimes = dimes
self.quarters = quarters
def penny_value(self):
return self.pennies * .01
def nickle_value(self):
return self.nickles * .05
def dime_value(self):
return self.dimes * .1
def quarter_value(self):
return self.quarters * .25
def add_coins(self):
return self.penny_value + self.nickle_value + self.dime_value + self.quarter_value
my_coins = Currency(1, 1, 1, 1)
#why does this work?
print(my_coins.penny_value() + my_coins.nickle_value())
#but not this?
print(my_coins.add_coins())
Because you are summing functions; Here is what you want to do:
def add_coins(self):
return self.penny_value() + self.nickle_value() + self.dime_value() + self.quarter_value()
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What is the correct way to deal with return of function in try block?
Option 1:
def someFnc1() -> bool:
try:
fetchSomeData()
return True
except:
return False
Option 2:
def someFnc2() -> bool:
try:
fetchSomeData()
except:
return False
return True
This is described in part in PEP-8:
Additionally, for all try/except clauses, limit the try clause to the absolute minimum amount of code necessary. Again, this avoids masking bugs.
Yes:
try:
value = collection[key]
except KeyError:
return key_not_found(key)
else:
return handle_value(value)
No:
try:
# Too broad!
return handle_value(collection[key])
except KeyError:
# Will also catch KeyError raised by handle_value()
return key_not_found(key)
Based on this your second version should be considered most pythonic (i.e. the absolute minimum of code inside the try clause).