Usage

The tokenize module has several quirks which make it complicated to work with (in my opinion, more complicated than necessary, but it is what it is).

The primary motivation of this guide is to document these quirks and behaviors, as such a document would have been very helpful to me when I first started using the module. Most of these behaviors were learned from experimentation and reading the source code. I have no idea what behaviors can be considered API guarantees, that is, the CPython developers may decide to change them in future Python versions. With that being said, the CPython developers are generally very conservative about changes to the standard library that might break downstream code, even for major releases. I will try to keep this guide updated as new Python versions are released. Issue reports and pull requests are most welcome.

Calling Syntax

The first thing you’ll notice when using tokenize() is that its calling API is rather odd. It does not accept a string. It does not accept a file-like object either. Rather, it requires the readline method of a bytes-mode file-like object. The bytes-mode part is important. If a file is opened in text mode ('r' instead of 'br'), tokenize() will fail with an exception:

>>> import tokenize
>>> with open('example.py') as f: # Incorrect, the default mode is 'r', not 'br'
...     for tok in tokenize.tokenize(f.readline):
...         print(tok)
Traceback (most recent call last):
  ...
TypeError: startswith first arg must be str or a tuple of str, not bytes
>>> with open('example.py', 'br') as f:
...     for tok in tokenize.tokenize(f.readline):
...         print(tok)
TokenInfo(type=62 (ENCODING), string='utf-8', start=(0, 0), end=(0, 0), line='')
TokenInfo(type=60 (COMMENT), string='# This is a an example file to be tokenized', start=(1, 0), end=(1, 43), line='# This is a an example file to be tokenized\n')
TokenInfo(type=61 (NL), string='\n', start=(1, 43), end=(1, 44), line='# This is a an example file to be tokenized\n')
TokenInfo(type=61 (NL), string='\n', start=(2, 0), end=(2, 1), line='\n')
TokenInfo(type=1 (NAME), string='def', start=(3, 0), end=(3, 3), line='def two():\n')
TokenInfo(type=1 (NAME), string='two', start=(3, 4), end=(3, 7), line='def two():\n')
TokenInfo(type=54 (OP), string='(', start=(3, 7), end=(3, 8), line='def two():\n')
TokenInfo(type=54 (OP), string=')', start=(3, 8), end=(3, 9), line='def two():\n')
TokenInfo(type=54 (OP), string=':', start=(3, 9), end=(3, 10), line='def two():\n')
TokenInfo(type=4 (NEWLINE), string='\n', start=(3, 10), end=(3, 11), line='def two():\n')
TokenInfo(type=5 (INDENT), string='    ', start=(4, 0), end=(4, 4), line='    return 1 + 1\n')
TokenInfo(type=1 (NAME), string='return', start=(4, 4), end=(4, 10), line='    return 1 + 1\n')
TokenInfo(type=2 (NUMBER), string='1', start=(4, 11), end=(4, 12), line='    return 1 + 1\n')
TokenInfo(type=54 (OP), string='+', start=(4, 13), end=(4, 14), line='    return 1 + 1\n')
TokenInfo(type=2 (NUMBER), string='1', start=(4, 15), end=(4, 16), line='    return 1 + 1\n')
TokenInfo(type=4 (NEWLINE), string='\n', start=(4, 16), end=(4, 17), line='    return 1 + 1\n')
TokenInfo(type=6 (DEDENT), string='', start=(5, 0), end=(5, 0), line='')
TokenInfo(type=0 (ENDMARKER), string='', start=(5, 0), end=(5, 0), line='')

To tokenize a string, you must encode it into a bytes, create an io.BytesIO object (not StringIO), and use the readline method of that object. If you find yourself doing this often, it may be useful to define a helper function.

>>> import io
>>> def tokenize_string(s):
...     return tokenize.tokenize(io.BytesIO(s.encode('utf-8')).readline)
>>> for tok in tokenize_string('hello + tokenize\n'):
...     print(tok)
TokenInfo(type=62 (ENCODING), string='utf-8', start=(0, 0), end=(0, 0), line='')
TokenInfo(type=1 (NAME), string='hello', start=(1, 0), end=(1, 5), line='hello + tokenize\n')
TokenInfo(type=54 (OP), string='+', start=(1, 6), end=(1, 7), line='hello + tokenize\n')
TokenInfo(type=1 (NAME), string='tokenize', start=(1, 8), end=(1, 16), line='hello + tokenize\n')
TokenInfo(type=4 (NEWLINE), string='\n', start=(1, 16), end=(1, 17), line='hello + tokenize\n')
TokenInfo(type=0 (ENDMARKER), string='', start=(2, 0), end=(2, 0), line='')

The reason for this API is that tokenize() is a streaming API, which works line-by-line on a Python document. This is also why the tokenize() function returns a generator. The typical pattern when using tokenize() is to iterate over it with a for loop (see the next section). If you are finished doing whatever you are doing before you reach the ENDMARKER token, you can break early from the loop efficiently without tokenizing the rest of the document. It is not recommended to convert the tokenize() generator into a list, except for debugging purposes. Not only is this inefficient, it makes it impossible to deal with exceptions.

TokenInfo

The tokenize() generator yields TokenInfo namedtuple objects, with the following fields:

>>> tokenize.TokenInfo._fields
('type', 'string', 'start', 'end', 'line')

The meaning of each field is outlined below.

There are two ways to work with TokenInfo objects. One is to unpack the tuple, typically in the for statement:

>>> for toknum, tokval, start, end, line in tokenize_string('hello + tokenize\n'):
...     print("toknum:", toknum, "tokval:", repr(tokval), "start:", start, "end:", end, "line:", repr(line))
toknum: 62 tokval: 'utf-8' start: (0, 0) end: (0, 0) line: ''
toknum: 1 tokval: 'hello' start: (1, 0) end: (1, 5) line: 'hello + tokenize\n'
toknum: 54 tokval: '+' start: (1, 6) end: (1, 7) line: 'hello + tokenize\n'
toknum: 1 tokval: 'tokenize' start: (1, 8) end: (1, 16) line: 'hello + tokenize\n'
toknum: 4 tokval: '\n' start: (1, 16) end: (1, 17) line: 'hello + tokenize\n'
toknum: 0 tokval: '' start: (2, 0) end: (2, 0) line: ''

By tradition, unused values are often replaced by _. You can also unpack the start and end tuples directly.

>>> for _, tokval, (start_line, start_col), (end_line, end_col), _ in tokenize_string('hello + tokenize\n'):
...     print("{tokval!r} on lines {start_line} to {end_line} on columns {start_col} to {end_col}".format(tokval=tokval, start_line=start_line, end_line=end_line, start_col=start_col, end_col=end_col))
'utf-8' on lines 0 to 0 on columns 0 to 0
'hello' on lines 1 to 1 on columns 0 to 5
'+' on lines 1 to 1 on columns 6 to 7
'tokenize' on lines 1 to 1 on columns 8 to 16
'\n' on lines 1 to 1 on columns 16 to 17
'' on lines 2 to 2 on columns 0 to 0

The other is to use it as-is, and access the members via attributes. I like using tok as the variable name for the tokens. token can be confused with the module name, so I don’t recommend using that (even though I recommend only importing tokenize, which includes all the names from token).

>>> for tok in tokenize_string('hello + tokenize\n'):
...     print("type:", tok.type, "string:", repr(tok.string), "start:", tok.start, "end:", tok.end, "line:", repr(tok.line))
type: 62 string: 'utf-8' start: (0, 0) end: (0, 0) line: ''
type: 1 string: 'hello' start: (1, 0) end: (1, 5) line: 'hello + tokenize\n'
type: 54 string: '+' start: (1, 6) end: (1, 7) line: 'hello + tokenize\n'
type: 1 string: 'tokenize' start: (1, 8) end: (1, 16) line: 'hello + tokenize\n'
type: 4 string: '\n' start: (1, 16) end: (1, 17) line: 'hello + tokenize\n'
type: 0 string: '' start: (2, 0) end: (2, 0) line: ''

One advantage of this second way is that the TokenInfo object contains an additional attribute, exact_type, which contains the exact token type of an OP token. However, this can also be determined from the string. Additionally, the first form is less verbose, but the second form avoids errors from getting the attributes in the wrong order. The form that you should use depends on your preference on these tradeoffs. I personally recommend the second form (for tok in tokenize(...): ... tok.type, etc.), unless you have the (toknum, tokstr, start, end, line) order memorized.

TokenInfo Fields

type

The token types are outlined in detail in the Token Types section.

string

The chunk of code that is tokenized. For token types where the string is meaningless, such as ENDMARKER, the string is empty.

For the ENCODING token, the string is the encoding, which does not appear literally in the code, which is why for ENCODING the line and column numbers are 0 and the line is the empty string.

start and end

start and end are tuples of (line number, column number) for the line and column numbers of the start and end of the tokenized string. Line numbers start at 1 and column numbers start at 0. The line and column numbers for the ENCODING token, which is always the first token emitted, are both (0, 0).

Because Python tuples compare lexicographically (i.e., (a, b) < (c, d) is equivalent to a < c or (a == c and b <= d)), these tuples can be compared directly to determine which comes earlier in the input. The start and end tuples as emitted from tokenize() are always nondecreasing (that is, start <= end will always be True for a single TokenInfo, and tok0.start <= tok1.start and tok0.end <= tok1.end will be True for consecutive TokenInfos, tok0 and tok1).

You should always use the start and end tuples to deduce line or column information. tokenize() ignores syntactically irrelevant whitespace, which can include newlines (in particular, escaped newlines, see NL).

line

line gives the full line that the token comes from. This is useful for reconstructing the whitespace between tokens (never assume that the whitespace between tokens is space characters—it could also consist of escaped newlines or tabs). line can also be useful for providing contextual error messages relating to the tokenization.

Exceptions

tokenize() has two failure modes: ERRORTOKEN and exceptions. When a non-fatal error occurs, some text will be tokenized as an ERRORTOKEN, and tokenizing will continue on the remainder of the input. This happens, for instance, for unrecognized characters, such as $, and unclosed single-quoted strings. See the ERRORTOKEN and STRING references for more information.

Other failures are so fatal that tokenization cannot continue, causing an exception to be raised. Depending on what you are doing, you may want to catch the exception and deal with it or let it bubble up to the caller.

These are the exceptions that can be raised from tokenize(). An exception other than these likely indicates incorrect input.

SyntaxError

SyntaxError is raised when the input has an invalid encoding. The encoding is detected using the detect_encoding() function, which uses either a PEP 263 formatted comment at the beginning of the input (like # -*- coding: utf-8 -*-), or a Unicode BOM character. If both are present, or if the PEP 263 encoding is invalid, SyntaxError is raised.

>>> tokenize.tokenize(io.BytesIO(b"# -*- coding: invalid -*-\n").readline)
Traceback (most recent call last):
  ...
SyntaxError: unknown encoding: invalid

See also ENCODING.

TokenError

tokenize.TokenError is raised in two situations. The only way to distinguish the two is to inspect the exception message. In both cases, TokenError is raised if the end of the input (EOF) is reached without a delimiter being closed. Tokens before the end of the input are still emitted, so it is typically desirable to process the tokens from tokenize() and handle TokenError as an end of input condition.

The second argument of the exception (e.args[1]) is a tuple of the start line and column number for the part of the input that was not tokenized due to the exception.

  1. EOF in multi-line string

    This happens if a triple-quoted string (i.e., a multi-line string, or “docstring”), is not closed at the end of the input. This exception can be detected by checking 'string' in e.args[0].

    >>> for tok in tokenize.tokenize(io.BytesIO(b"""
    ... def f():
    ...     '''
    ...     unclosed docstring
    ... """).readline):
    ...     print(tok) 
    TokenInfo(type=62 (ENCODING), string='utf-8', start=(0, 0), end=(0, 0), line='')
    TokenInfo(type=61 (NL), string='\n', start=(1, 0), end=(1, 1), line='\n')
    TokenInfo(type=1 (NAME), string='def', start=(2, 0), end=(2, 3), line='def f():\n')
    TokenInfo(type=1 (NAME), string='f', start=(2, 4), end=(2, 5), line='def f():\n')
    TokenInfo(type=54 (OP), string='(', start=(2, 5), end=(2, 6), line='def f():\n')
    TokenInfo(type=54 (OP), string=')', start=(2, 6), end=(2, 7), line='def f():\n')
    TokenInfo(type=54 (OP), string=':', start=(2, 7), end=(2, 8), line='def f():\n')
    TokenInfo(type=4 (NEWLINE), string='\n', start=(2, 8), end=(2, 9), line='def f():\n')
    TokenInfo(type=5 (INDENT), string='    ', start=(3, 0), end=(3, 4), line="    '''\n")
    Traceback (most recent call last):
      ...
    tokenize.TokenError: ('EOF in multi-line string', (3, 4))
    
  2. EOF in multi-line statement

    This error occurs when an unclosed brace is found. Note that tokenize does not necessarily stop parsing as soon as the input is syntactically invalid, as it only has limited knowledge of Python syntax. In fact, in the current implementation, this exception is only raised after all tokens have been emitted, except possibly DEDENT tokens. This exception can be detected by checking 'statement' in e.args[0].

    >>> for tok in tokenize.tokenize(io.BytesIO(b"""
    ... (1 +
    ... def f():
    ...     pass
    ... """).readline):
    ...     print(tok) 
    TokenInfo(type=62 (ENCODING), string='utf-8', start=(0, 0), end=(0, 0), line='')
    TokenInfo(type=61 (NL), string='\n', start=(1, 0), end=(1, 1), line='\n')
    TokenInfo(type=54 (OP), string='(', start=(2, 0), end=(2, 1), line='(1 +\n')
    TokenInfo(type=2 (NUMBER), string='1', start=(2, 1), end=(2, 2), line='(1 +\n')
    TokenInfo(type=54 (OP), string='+', start=(2, 3), end=(2, 4), line='(1 +\n')
    TokenInfo(type=61 (NL), string='\n', start=(2, 4), end=(2, 5), line='(1 +\n')
    TokenInfo(type=1 (NAME), string='def', start=(3, 0), end=(3, 3), line='def f():\n')
    TokenInfo(type=1 (NAME), string='f', start=(3, 4), end=(3, 5), line='def f():\n')
    TokenInfo(type=54 (OP), string='(', start=(3, 5), end=(3, 6), line='def f():\n')
    TokenInfo(type=54 (OP), string=')', start=(3, 6), end=(3, 7), line='def f():\n')
    TokenInfo(type=54 (OP), string=':', start=(3, 7), end=(3, 8), line='def f():\n')
    TokenInfo(type=61 (NL), string='\n', start=(3, 8), end=(3, 9), line='def f():\n')
    TokenInfo(type=1 (NAME), string='pass', start=(4, 4), end=(4, 8), line='    pass\n')
    TokenInfo(type=61 (NL), string='\n', start=(4, 8), end=(4, 9), line='    pass\n')
    Traceback (most recent call last):
      ...
    tokenize.TokenError: ('EOF in multi-line statement', (5, 0))
    

The following example shows how TokenError might be caught and processed. See also the examples.

>>> def tokens_with_errors(s):
...     try:
...         for tok in tokenize.tokenize(io.BytesIO(s.encode('utf-8')).readline):
...             print(tok)
...     except tokenize.TokenError as e:
...         if 'string' in e.args[0]:
...             print("TokenError: Unclosed multi-line string starting at", e.args[1])
...         elif 'statement' in e.args[0]:
...             print("TokenError: Unclosed brace(s)")
...         else:
...             # Unrecognized TokenError. Shouldn't happen
...             raise
>>> tokens_with_errors("""
... def f():
...     '''
...     unclosed docstring
... """)
TokenInfo(type=62 (ENCODING), string='utf-8', start=(0, 0), end=(0, 0), line='')
TokenInfo(type=61 (NL), string='\n', start=(1, 0), end=(1, 1), line='\n')
TokenInfo(type=1 (NAME), string='def', start=(2, 0), end=(2, 3), line='def f():\n')
TokenInfo(type=1 (NAME), string='f', start=(2, 4), end=(2, 5), line='def f():\n')
TokenInfo(type=54 (OP), string='(', start=(2, 5), end=(2, 6), line='def f():\n')
TokenInfo(type=54 (OP), string=')', start=(2, 6), end=(2, 7), line='def f():\n')
TokenInfo(type=54 (OP), string=':', start=(2, 7), end=(2, 8), line='def f():\n')
TokenInfo(type=4 (NEWLINE), string='\n', start=(2, 8), end=(2, 9), line='def f():\n')
TokenInfo(type=5 (INDENT), string='    ', start=(3, 0), end=(3, 4), line="    '''\n")
TokenError: Unclosed multi-line string starting at (3, 4)
>>> tokens_with_errors("""
... (1 +
... def f():
...     pass
... """)
TokenInfo(type=62 (ENCODING), string='utf-8', start=(0, 0), end=(0, 0), line='')
TokenInfo(type=61 (NL), string='\n', start=(1, 0), end=(1, 1), line='\n')
TokenInfo(type=54 (OP), string='(', start=(2, 0), end=(2, 1), line='(1 +\n')
TokenInfo(type=2 (NUMBER), string='1', start=(2, 1), end=(2, 2), line='(1 +\n')
TokenInfo(type=54 (OP), string='+', start=(2, 3), end=(2, 4), line='(1 +\n')
TokenInfo(type=61 (NL), string='\n', start=(2, 4), end=(2, 5), line='(1 +\n')
TokenInfo(type=1 (NAME), string='def', start=(3, 0), end=(3, 3), line='def f():\n')
TokenInfo(type=1 (NAME), string='f', start=(3, 4), end=(3, 5), line='def f():\n')
TokenInfo(type=54 (OP), string='(', start=(3, 5), end=(3, 6), line='def f():\n')
TokenInfo(type=54 (OP), string=')', start=(3, 6), end=(3, 7), line='def f():\n')
TokenInfo(type=54 (OP), string=':', start=(3, 7), end=(3, 8), line='def f():\n')
TokenInfo(type=61 (NL), string='\n', start=(3, 8), end=(3, 9), line='def f():\n')
TokenInfo(type=1 (NAME), string='pass', start=(4, 4), end=(4, 8), line='    pass\n')
TokenInfo(type=61 (NL), string='\n', start=(4, 8), end=(4, 9), line='    pass\n')
TokenError: Unclosed brace(s)

IndentationError

tokenize() raises IndentationError if an unindent does not match an outer indentation level.

>>> for tok in tokenize.tokenize(io.BytesIO(b"""
... if x:
...     pass
...  f
... """).readline):
...     print(tok) 
TokenInfo(type=62 (ENCODING), string='utf-8', start=(0, 0), end=(0, 0), line='')
TokenInfo(type=61 (NL), string='\n', start=(1, 0), end=(1, 1), line='\n')
TokenInfo(type=1 (NAME), string='if', start=(2, 0), end=(2, 2), line='if x:\n')
TokenInfo(type=1 (NAME), string='x', start=(2, 3), end=(2, 4), line='if x:\n')
TokenInfo(type=54 (OP), string=':', start=(2, 4), end=(2, 5), line='if x:\n')
TokenInfo(type=4 (NEWLINE), string='\n', start=(2, 5), end=(2, 6), line='if x:\n')
TokenInfo(type=5 (INDENT), string='    ', start=(3, 0), end=(3, 4), line='    pass\n')
TokenInfo(type=1 (NAME), string='pass', start=(3, 4), end=(3, 8), line='    pass\n')
TokenInfo(type=4 (NEWLINE), string='\n', start=(3, 8), end=(3, 9), line='    pass\n')
  File "<tokenize>", line 4
    f
    ^
IndentationError: unindent does not match any outer indentation level

This error is difficult to recover from. If you need to handle tokenizing input with invalid indentation, my best recommendation is to instead use the parso library, which does not raise IndentationError (it also does not raise any of the other exceptions discussed here). See also the discussion of parso in the alternatives section.

This is the only indentation error tokenize cares about. It does not care about other syntactically invalid constructs such as inconsistently mixing tabs and spaces.

>>> for tok in tokenize.tokenize(io.BytesIO(b"""
... if x:
...     \tpass
... \t    pass
... """).readline):
...     print(tok)
TokenInfo(type=62 (ENCODING), string='utf-8', start=(0, 0), end=(0, 0), line='')
TokenInfo(type=61 (NL), string='\n', start=(1, 0), end=(1, 1), line='\n')
TokenInfo(type=1 (NAME), string='if', start=(2, 0), end=(2, 2), line='if x:\n')
TokenInfo(type=1 (NAME), string='x', start=(2, 3), end=(2, 4), line='if x:\n')
TokenInfo(type=54 (OP), string=':', start=(2, 4), end=(2, 5), line='if x:\n')
TokenInfo(type=4 (NEWLINE), string='\n', start=(2, 5), end=(2, 6), line='if x:\n')
TokenInfo(type=5 (INDENT), string='    \t', start=(3, 0), end=(3, 5), line='    \tpass\n')
TokenInfo(type=1 (NAME), string='pass', start=(3, 5), end=(3, 9), line='    \tpass\n')
TokenInfo(type=4 (NEWLINE), string='\n', start=(3, 9), end=(3, 10), line='    \tpass\n')
TokenInfo(type=5 (INDENT), string='\t    ', start=(4, 0), end=(4, 5), line='\t    pass\n')
TokenInfo(type=1 (NAME), string='pass', start=(4, 5), end=(4, 9), line='\t    pass\n')
TokenInfo(type=4 (NEWLINE), string='\n', start=(4, 9), end=(4, 10), line='\t    pass\n')
TokenInfo(type=6 (DEDENT), string='', start=(5, 0), end=(5, 0), line='')
TokenInfo(type=6 (DEDENT), string='', start=(5, 0), end=(5, 0), line='')
TokenInfo(type=0 (ENDMARKER), string='', start=(5, 0), end=(5, 0), line='')
>>> exec("""
... if x:
...     \tpass
... \t    pass
... """)
  File "<string>", line 4
    pass
       ^
TabError: inconsistent use of tabs and spaces in indentation