pandas.Series.replace — pandas 2.2.3 documentation (original) (raw)

Series.replace(to_replace=None, value=<no_default>, *, inplace=False, limit=None, regex=False, method=<no_default>)[source]#

Replace values given in to_replace with value.

Values of the Series/DataFrame 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_replacestr, 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 withvalue

• list of str, regex, or numeric:

  • First, if to_replace and value are both lists, theymust be the same length.
  • Second, if regex=True then all of the strings in bothlists 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 valueparameter 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 valueparameter 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.

valuescalar, 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.

inplacebool, default False

If True, performs operation inplace and returns None.

limitint, default None

Maximum size gap to forward or backward fill.

Deprecated since version 2.1.0.

regexbool or same types as to_replace, default False

Whether to interpret to_replace and/or value as regular expressions. Alternatively, this could be a regular expression or a list, dict, or array of regular expressions in which caseto_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 since version 2.1.0.

Returns:

Series/DataFrame

Object after replacement.

Raises:

AssertionError

• If regex is not a bool and to_replace is notNone.

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 andvalue but they are not the same length.

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}) ands.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 tos.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 since version 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

Changed in version 1.4.0: Previously the explicit None was silently ignored.

When regex=True, value is not None and to_replace is a string, the replacement will be applied in all columns of the DataFrame.

df = pd.DataFrame({'A': [0, 1, 2, 3, 4], ... 'B': ['a', 'b', 'c', 'd', 'e'], ... 'C': ['f', 'g', 'h', 'i', 'j']})

df.replace(to_replace='^[a-g]', value='e', regex=True) A B C 0 0 e e 1 1 e e 2 2 e h 3 3 e i 4 4 e j

If value is not None and to_replace is a dictionary, the dictionary keys will be the DataFrame columns that the replacement will be applied.

df.replace(to_replace={'B': '^[a-c]', 'C': '^[h-j]'}, value='e', regex=True) A B C 0 0 e f 1 1 e g 2 2 e e 3 3 d e 4 4 e e