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# pandas.DataFrame.query

`DataFrame.``query`(expr, **kwargs)

  Query the columns of a frame with a boolean expression.

  New in version 0.13.

  Parameters : | expr : string The query string to evaluate. You can refer to variables in the environment by prefixing them with an ‘@’ character like `@a + b`. kwargs : dict See the documentation for `pandas.eval()` for complete details on the keyword arguments accepted by `DataFrame.query()`.
  Returns : | q : DataFrame

  See also

  `pandas.eval`, `DataFrame.eval`

  Notes

  The result of the evaluation of this expression is first passed to `DataFrame.loc` and if that fails because of a multidimensional key (e.g., a DataFrame) then the result will be passed to `DataFrame.__getitem__()`.

  This method uses the top-level `pandas.eval()` function to evaluate the passed query.

  The `query()` method uses a slightly modified Python syntax by default. For example, the `&` and `|` (bitwise) operators have the precedence of their boolean cousins, `and` and `or`. This is syntactically valid Python, however the semantics are different.

  You can change the semantics of the expression by passing the keyword argument `parser='python'`. This enforces the same semantics as evaluation in Python space. Likewise, you can pass `engine='python'` to evaluate an expression using Python itself as a backend. This is not recommended as it is inefficient compared to using `numexpr` as the engine.

  The `DataFrame.index` and `DataFrame.columns` attributes of the `DataFrame` instance are placed in the query namespace by default, which allows you to treat both the index and columns of the frame as a column in the frame. The identifier `index` is used for the frame index; you can also use the name of the index to identify it in a query.

  For further details and examples see the `query` documentation in indexing.

  Examples

      >>> from numpy.random import randn
      >>> from pandas import DataFrame
      >>> df = DataFrame(randn(10, 2), columns=list('ab'))
      >>> df.query('a > b')
      >>> df[df.a > df.b]  # same result as the previous expression

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