ue` * 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 optional `value` parameter should not be given. - 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 optional `value` parameter should not be specified 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} limit : int, default None Maximum size gap to forward or backward fill. .. deprecated:: 2.1.0 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'}} The method to use when for replacement, when `to_replace` is a scalar, list or tuple and `value` is ``None``. .. deprecated:: 2.1.0 Returns ------- {klass} Object after replacement. 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. DataFrame.fillna : Fill NA values. Series.where : Replace values based on boolean condition. DataFrame.where : Replace values based on boolean condition. DataFrame.map: Apply a function to a Dataframe elementwise. Series.map: Map values of Series according to an input mapping or function. 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([1, 2, 3, 4, 5]) >>> s.replace(1, 5) 0 5 1 2 2 3 3 4 4 5 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 3 1 3 2 3 3 4 4 5 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`` is not explicitly passed 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. >>> s.replace('a') 0 10 1 10 2 10 3 b 4 b dtype: object .. deprecated:: 2.1.0 The 'method' parameter and padding behavior are deprecated. On the other hand, if ``None`` is explicitly passed for ``value``, it will be respected: >>> s.replace('a', None) 0 10 1 None 2 None 3 b 4 None dtype: object .. versionchanged:: 1.4.0 Previously the explicit ``None`` was silently ignored. Ú