Some common errors in python

Common mistakes in python:

0. Forgot to write colon

in if, elif, else, for, while, class, def statements Forgetting to add ":"

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

result in: SyntaxError: invalid syntax

2. Using wrong indentation

Python uses indentation to distinguish code blocks, common incorrect usages :

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

Causes: IndentationError: unexpected indent. Each line of code in the same code block must maintain a consistent indentation

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

Cause: IndentationError: unindent does not match any outer indentation level. After the code block ends, the indentation returns to its original position

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

Results in: IndentationError: expected an indented block, ":" needs to be followed by indentation

3. The variable is not defined

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

Results in: NameError: name 'spam' is not defined

4. When obtaining the index position of the list element, forget to call the len method

When obtaining the element through the index position, forget to use the len function to obtain it The length of the list.

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

Results in: TypeError: range() integer end argument expected, got list.
The correct approach is:

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

Of course, the more Pythonic way is to use enumerate

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

5. Modify the string

The string is a sequence object that supports obtaining elements by index, but it is different from the list object. The string is an immutable object and does not support modification.

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

Results in: TypeError: 'str' object does not support item assignment

The correct approach should be:

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

6. String and non-string concatenation

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

Results in: TypeError: cannot concatenate 'str' and 'int' objects

When strings are connected to non-strings, the non-string objects must be coerced into string types

num_eggs = 12
print('I have ' + str(num_eggs) + ' eggs.')

Or use the formatting form of the string

num_eggs = 12
print('I have %s eggs.' % (num_eggs))

7. Using the wrong index position

spam = ['cat', 'dog', 'mouse']
print(spam[3])

results in: IndexError: list index out of range

of the list object The index starts from 0, and the third element should be accessed using spam[2]

8. Using non-existent keys in the dictionary

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

To access keys in the dictionary object, you can use [ ], but if the key does not exist, it will result in: KeyError: 'zebra'

The correct way should be to use the get method

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

When the key does not exist, get returns None by default

9. Using keywords as variable names

class = 'algebra'

Results in: SyntaxError: invalid syntax

It is not allowed to use keywords as variable names in Python. Python3 has a total of 33 keywords.

>>> import keyword
>>> print(keyword.kwlist)
['False', 'None', 'True', 'and', 'as', 'assert', 'break', 'class', 'continue', 'def', 'del', 'elif', 'else', 'except', 'finally', 'for', 'from', 'global', 'if', 'import', 'in', 'is', 'lambda', 'nonlocal', 'not', 'or', 'pass', 'raise', 'return', 'try', 'while', 'with', 'yield']

10. The local variable in the function is used before assignment

someVar = 42

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

myFunction()

Results in: UnboundLocalError: local variable 'someVar' referenced before assignment

When there is a global variable in the function When there is a variable with the same name in the domain, it will search for the variable in LEGB order. If a variable with the same name is also defined in the local scope inside the function, it will not search in the external scope.

Therefore, someVar is defined in the myFunction function, so print(someVar) no longer searches outside, but the variable has not been assigned a value when printing, so UnboundLocalError

11 occurs , use auto-increment " " auto-decrement "--"

spam = 0
spam++

Haha, there are no auto-increment and self-decrement operators in Python. If you are transferring from C or Java, you should pay attention. You can use " = " instead of " "

#method1 is a member method of the Foo class. This method does not accept any parameters. Calling a.method1() is equivalent to calling Foo.method1(a), but method1 does not accept any parameters, so an error is reported. The correct calling method should be Foo.method1().

For more related knowledge, please pay attention to

python video tutorial

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