Classe « Series »
Signature de la méthode replace
def replace(self, to_replace=None, value=None, inplace=False, limit=None, regex=False, method='pad')
Description
replace.__doc__
Replace values given in `to_replace` with `value`.
Values of the Series are replaced with other values dynamically.
This differs from updating with ``.loc`` or ``.iloc``, which require
you to specify a location to update with some value.
Parameters
----------
to_replace : str, regex, list, dict, Series, int, float, or None
How to find the values that will be replaced.
* numeric, str or regex:
- numeric: numeric values equal to `to_replace` will be
replaced with `value`
- str: string exactly matching `to_replace` will be replaced
with `value`
- regex: regexs matching `to_replace` will be replaced with
`value`
* list of str, regex, or numeric:
- First, if `to_replace` and `value` are both lists, they
**must** be the same length.
- Second, if ``regex=True`` then all of the strings in **both**
lists will be interpreted as regexs otherwise they will match
directly. This doesn't matter much for `value` since there
are only a few possible substitution regexes you can use.
- str, regex and numeric rules apply as above.
* dict:
- Dicts can be used to specify different replacement values
for different existing values. For example,
``{'a': 'b', 'y': 'z'}`` replaces the value 'a' with 'b' and
'y' with 'z'. To use a dict in this way the `value`
parameter should be `None`.
- For a DataFrame a dict can specify that different values
should be replaced in different columns. For example,
``{'a': 1, 'b': 'z'}`` looks for the value 1 in column 'a'
and the value 'z' in column 'b' and replaces these values
with whatever is specified in `value`. The `value` parameter
should not be ``None`` in this case. You can treat this as a
special case of passing two lists except that you are
specifying the column to search in.
- For a DataFrame nested dictionaries, e.g.,
``{'a': {'b': np.nan}}``, are read as follows: look in column
'a' for the value 'b' and replace it with NaN. The `value`
parameter should be ``None`` to use a nested dict in this
way. You can nest regular expressions as well. Note that
column names (the top-level dictionary keys in a nested
dictionary) **cannot** be regular expressions.
* None:
- This means that the `regex` argument must be a string,
compiled regular expression, or list, dict, ndarray or
Series of such elements. If `value` is also ``None`` then
this **must** be a nested dictionary or Series.
See the examples section for examples of each of these.
value : scalar, dict, list, str, regex, default None
Value to replace any values matching `to_replace` with.
For a DataFrame a dict of values can be used to specify which
value to use for each column (columns not in the dict will not be
filled). Regular expressions, strings and lists or dicts of such
objects are also allowed.
inplace : bool, default False
If True, in place. Note: this will modify any
other views on this object (e.g. a column from a DataFrame).
Returns the caller if this is True.
limit : int or None, default None
Maximum size gap to forward or backward fill.
regex : bool or same types as `to_replace`, default False
Whether to interpret `to_replace` and/or `value` as regular
expressions. If this is ``True`` then `to_replace` *must* be a
string. Alternatively, this could be a regular expression or a
list, dict, or array of regular expressions in which case
`to_replace` must be ``None``.
method : {'pad', 'ffill', 'bfill', `None`}
The method to use when for replacement, when `to_replace` is a
scalar, list or tuple and `value` is ``None``.
Returns
-------
Series or None
Object after replacement or None if ``inplace=True``.
Raises
------
AssertionError
* If `regex` is not a ``bool`` and `to_replace` is not
``None``.
TypeError
* If `to_replace` is not a scalar, array-like, ``dict``, or ``None``
* If `to_replace` is a ``dict`` and `value` is not a ``list``,
``dict``, ``ndarray``, or ``Series``
* If `to_replace` is ``None`` and `regex` is not compilable
into a regular expression or is a list, dict, ndarray, or
Series.
* When replacing multiple ``bool`` or ``datetime64`` objects and
the arguments to `to_replace` does not match the type of the
value being replaced
ValueError
* If a ``list`` or an ``ndarray`` is passed to `to_replace` and
`value` but they are not the same length.
See Also
--------
Series.fillna : Fill NA values.
Series.where : Replace values based on boolean condition.
Series.str.replace : Simple string replacement.
Notes
-----
* Regex substitution is performed under the hood with ``re.sub``. The
rules for substitution for ``re.sub`` are the same.
* Regular expressions will only substitute on strings, meaning you
cannot provide, for example, a regular expression matching floating
point numbers and expect the columns in your frame that have a
numeric dtype to be matched. However, if those floating point
numbers *are* strings, then you can do this.
* This method has *a lot* of options. You are encouraged to experiment
and play with this method to gain intuition about how it works.
* When dict is used as the `to_replace` value, it is like
key(s) in the dict are the to_replace part and
value(s) in the dict are the value parameter.
Examples
--------
**Scalar `to_replace` and `value`**
>>> s = pd.Series([0, 1, 2, 3, 4])
>>> s.replace(0, 5)
0 5
1 1
2 2
3 3
4 4
dtype: int64
>>> df = pd.DataFrame({'A': [0, 1, 2, 3, 4],
... 'B': [5, 6, 7, 8, 9],
... 'C': ['a', 'b', 'c', 'd', 'e']})
>>> df.replace(0, 5)
A B C
0 5 5 a
1 1 6 b
2 2 7 c
3 3 8 d
4 4 9 e
**List-like `to_replace`**
>>> df.replace([0, 1, 2, 3], 4)
A B C
0 4 5 a
1 4 6 b
2 4 7 c
3 4 8 d
4 4 9 e
>>> df.replace([0, 1, 2, 3], [4, 3, 2, 1])
A B C
0 4 5 a
1 3 6 b
2 2 7 c
3 1 8 d
4 4 9 e
>>> s.replace([1, 2], method='bfill')
0 0
1 3
2 3
3 3
4 4
dtype: int64
**dict-like `to_replace`**
>>> df.replace({0: 10, 1: 100})
A B C
0 10 5 a
1 100 6 b
2 2 7 c
3 3 8 d
4 4 9 e
>>> df.replace({'A': 0, 'B': 5}, 100)
A B C
0 100 100 a
1 1 6 b
2 2 7 c
3 3 8 d
4 4 9 e
>>> df.replace({'A': {0: 100, 4: 400}})
A B C
0 100 5 a
1 1 6 b
2 2 7 c
3 3 8 d
4 400 9 e
**Regular expression `to_replace`**
>>> df = pd.DataFrame({'A': ['bat', 'foo', 'bait'],
... 'B': ['abc', 'bar', 'xyz']})
>>> df.replace(to_replace=r'^ba.$', value='new', regex=True)
A B
0 new abc
1 foo new
2 bait xyz
>>> df.replace({'A': r'^ba.$'}, {'A': 'new'}, regex=True)
A B
0 new abc
1 foo bar
2 bait xyz
>>> df.replace(regex=r'^ba.$', value='new')
A B
0 new abc
1 foo new
2 bait xyz
>>> df.replace(regex={r'^ba.$': 'new', 'foo': 'xyz'})
A B
0 new abc
1 xyz new
2 bait xyz
>>> df.replace(regex=[r'^ba.$', 'foo'], value='new')
A B
0 new abc
1 new new
2 bait xyz
Compare the behavior of ``s.replace({'a': None})`` and
``s.replace('a', None)`` to understand the peculiarities
of the `to_replace` parameter:
>>> s = pd.Series([10, 'a', 'a', 'b', 'a'])
When one uses a dict as the `to_replace` value, it is like the
value(s) in the dict are equal to the `value` parameter.
``s.replace({'a': None})`` is equivalent to
``s.replace(to_replace={'a': None}, value=None, method=None)``:
>>> s.replace({'a': None})
0 10
1 None
2 None
3 b
4 None
dtype: object
When ``value=None`` and `to_replace` is a scalar, list or
tuple, `replace` uses the method parameter (default 'pad') to do the
replacement. So this is why the 'a' values are being replaced by 10
in rows 1 and 2 and 'b' in row 4 in this case.
The command ``s.replace('a', None)`` is actually equivalent to
``s.replace(to_replace='a', value=None, method='pad')``:
>>> s.replace('a', None)
0 10
1 10
2 10
3 b
4 b
dtype: object
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