Skip to main content
`Ctrl`+`K`

# pandas.DataFrame.replace

DataFrame.replace(to_replace=None, value=<no_default>, *, inplace=False, regex=False)

  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_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: regexes 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 regexes 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 bool, default False

      If True, performs operation inplace.

    regex bool 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 case to_replace must be `None`. Patterns may be passed either as strings or as compiled regex objects (`re.compile(...)`); use a compiled object when you need to set flags such as `re.IGNORECASE`, since `replace` does not accept a separate `flags` argument.

  Returns:
    Series/DataFrame

      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.

  - Unless regex is True, a key matches only if it equals the entire value, not a substring of it: `{"b": "z"}` replaces the value `"b"` but leaves `"a b c"` unchanged.

  - Dict entries are applied one after another, in insertion order. Which entries apply to an element is decided up front by matching every key against the original value, but each substitution is made into the result of the previous ones. An entry can therefore end up a no-op if an earlier one already rewrote the text it targeted: with `regex=True`, `{"x": "1", "x y": "2"}` turns `"x y"` into `"1 y"`, while `{"x y": "2", "x": "1"}` turns it into `"2"`. Conversely, a later entry can rewrite what an earlier one produced: `{"ab": "ba", "ba": "zz"}` turns `"ab ba"` into `"zz zz"`. With overlapping keys, order the dict most-specific-first.

  - Replacement is based on equality, not identity. Since Python treats `True == 1` and `False == 0`, replacing one will also affect the other when they share a dtype (e.g. `object`).

  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

  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

  To match case-insensitively (or apply any other `re` flag), pass a compiled regex object. `replace` does not take a separate `flags` argument, so the flags must be baked into the compiled pattern:

      >>> import re
      >>> df = pd.DataFrame({"A": ["Foo", "FOO", "bar"], "B": ["foo", "Bar", "BAR"]})
      >>> df.replace(re.compile(r"foo", flags=re.IGNORECASE), "new", regex=True)
           A    B
      0  new  new
      1  new  Bar
      2  bar  BAR

  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)`:

      >>> s.replace({"a": None})
      0      10
      1    None
      2    None
      3       b
      4    None
      dtype: object

  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

  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