Excellent! Finally, we have sorted out Python multithreading and multiprocessing!

Bald programmer 2022-06-23 17:54:01 阅读数:651

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For the newcomer Pythoner In the process of learning to run the code is more or less will encounter some errors , It may seem like a lot of work at first . As the amount of code accumulates , Practice makes perfect when encountering some runtime errors, it can quickly locate the original problem . Here are some common 17 A mistake , I hope I can help you .

1、

Forget in if,for,def,elif,else,class Wait until the end of the statement : It can lead to “SyntaxError :invalid syntax” as follows :

if spam == 42
print('Hello!')

2、

Use = instead of ==

It can also lead to “SyntaxError: invalid syntax”= It's an assignment operator and == It's a comparison operation . This error occurs in the following code :

if spam = 42:
print('Hello!')

3、

Incorrect use of indents leads to “IndentationError:unexpected indent”、“IndentationError:unindent does not match any outer indetation level” as well as “IndentationError:expected an indented block” Remember that indent increase is only used to : After the closing statement , And then you have to go back to the previous indent format . This error occurs in the following code :

print('Hello!')
print('Howdy!')

perhaps :

if spam == 42:
print('Hello!')
print('Howdy!')

4、

stay for Forget to call... In the loop statement len()

Lead to “TypeError: 'list' object cannot be interpreted as an integer

Usually you want to iterate through an index list perhaps string The elements of , This requires calling range() function . Remember to go back to len Value instead of returning this list .

This error occurs in the following code :

spam = ['cat', 'dog', 'mouse']
for i in range(spam):
print(spam[i])

5、 Try to modify string The value of “TypeError: 'str' object does not support item assignment”string It's an immutable data type , This error occurs in the following code :

spam = 'I have a pet cat.'
spam[13] = 'r'
print(spam)

And the right thing to do is :

spam = 'I have a pet cat.'
spam = spam[:13] + 'r' + spam[14:]
print(spam)

6、 Trying to connect a non string value to a string results in “TypeError: Can't convert 'int' object to str implicitly” This error occurs in the following code :

numEggs = 12
print('I have ' + numEggs + ' eggs.')

And the right thing to do is :

numEggs = 12
print('I have ' + str(numEggs) + ' eggs.')
numEggs = 12
print('I have %s eggs.' % (numEggs))

7、 Forgetting to put quotation marks at the beginning and end of a string leads to “SyntaxError: EOL while scanning string literal” This error occurs in the following code :

print(Hello!')
print('Hello!)
myName = 'Al'
print('My name is ' + myName + . How are you?')

8、

Misspelling variable or function names leads to “NameError: name 'fooba' is not defined” This error occurs in the following code :

foobar = 'Al'
print('My name is ' + fooba)
spam = ruond(4.2)
spam = Round(4.2)

9、 A misspelled method name results in “AttributeError: 'str' object has no attribute 'lowerr' ” This error occurs in the following code :

spam = 'THIS IS IN LOWERCASE.'
spam = spam.lowerr()

10、

Reference exceeds list Maximum index results in “IndexError: list index out of range” This error occurs in the following code :

spam = ['cat', 'dog', 'mouse']
print(spam[6])

11、 Using non-existent dictionary key values results in “KeyError:‘spam’” This error occurs in the following code :

spam = {'cat': 'Zophie', 'dog': 'Basil', 'mouse': 'Whiskers'}
print('The name of my pet zebra is ' + spam['zebra'])

12、 Try to use Python Keyword as variable name leads to “SyntaxError:invalid syntax”Python Key cannot be used as variable name , This error occurs in the following code :

class = 'algebra'

Python3 The key words are :and, as, assert, break, class, continue, def, del, elif, else, except, False, finally, for, from, global, if, import, in, is, lambda, None, nonlocal, not, or, pass, raise, return, True, try, while, with, yield

13、

Use the value-added operator... In a new variable definition

Lead to “NameError: name 'foobar' is not defined

Do not use... When declaring variables 0 Or an empty string as the initial value , In this way, we use a sentence of the autoincrement operator spam += 1 be equal to spam = spam + 1, It means spam You need to specify a valid initial value .

This error occurs in the following code :

spam = 0
spam += 42
eggs += 42

14、 Use local variables in functions before defining local variables ( At this time, there is a global variable with the same name as the local variable ) Lead to “UnboundLocalError: local variable 'foobar' referenced before assignment” It's very complicated to use local variables in a function and have a global variable with the same name at the same time , The rule of use is : If anything is defined in a function , If it's only used in functions, then it's local , On the contrary, it's a global variable . This means that you can't use it as a global variable in a function before defining it . This error occurs in the following code :

someVar = 42
def myFunction():
print(someVar)
someVar = 100
myFunction()

15、 Try to use range() Creating a list of integers results in “TypeError: 'range' object does not support item assignment” Sometimes you want to get an ordered list of integers , therefore range() Looks like a good way to generate this list . However , You need to remember range() The return is “range object”, Not the actual list value . This error occurs in the following code :

spam = range(10)
spam[4] = -1

Write it correctly :

spam = list(range(10))
spam[4] = -1

( Be careful : stay Python 2 in spam = range(10) It can be done , Because in Python 2 in range() The return is list value , But in Python 3 The above mistakes will occur in )

16、 non-existent ++ perhaps -- Self - increment and self - decrement operators . Lead to “SyntaxError: invalid syntax” If you are used to, for example C++ , Java , PHP Wait for other languages , Maybe you want to try to use ++ perhaps -- To increase or decrease by one variable . stay Python There is no such operator in . This error occurs in the following code :

spam = 1
spam++

Write it correctly :

spam = 1
spam += 1

17、 Forget to add... For the first parameter of the method self Parameter result “TypeError: myMethod() takes no arguments (1 given) ” This error occurs in the following code :

class Foo():
def myMethod():
print('Hello!')
a = Foo()
a.myMethod()
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