r"""
    Wide panel to long format. Less flexible but more user-friendly than melt.

    With stubnames ['A', 'B'], this function expects to find one or more
    group of columns with format
    A-suffix1, A-suffix2,..., B-suffix1, B-suffix2,...
    You specify what you want to call this suffix in the resulting long format
    with `j` (for example `j='year'`)

    Each row of these wide variables are assumed to be uniquely identified by
    `i` (can be a single column name or a list of column names)

    All remaining variables in the data frame are left intact.

    Parameters
    ----------
    df : DataFrame
        The wide-format DataFrame
    stubnames : str or list-like
        The stub name(s). The wide format variables are assumed to
        start with the stub names.
    i : str or list-like
        Column(s) to use as id variable(s)
    j : str
        The name of the sub-observation variable. What you wish to name your
        suffix in the long format.
    sep : str, default ""
        A character indicating the separation of the variable names
        in the wide format, to be stripped from the names in the long format.
        For example, if your column names are A-suffix1, A-suffix2, you
        can strip the hyphen by specifying `sep='-'`

        .. versionadded:: 0.20.0

    suffix : str, default '\\d+'
        A regular expression capturing the wanted suffixes. '\\d+' captures
        numeric suffixes. Suffixes with no numbers could be specified with the
        negated character class '\\D+'. You can also further disambiguate
        suffixes, for example, if your wide variables are of the form
        A-one, B-two,.., and you have an unrelated column A-rating, you can
        ignore the last one by specifying `suffix='(!?one|two)'`

        .. versionadded:: 0.20.0

        .. versionchanged:: 0.23.0
            When all suffixes are numeric, they are cast to int64/float64.

    Returns
    -------
    DataFrame
        A DataFrame that contains each stub name as a variable, with new index
        (i, j).

    Notes
    -----
    All extra variables are left untouched. This simply uses
    `pandas.melt` under the hood, but is hard-coded to "do the right thing"
    in a typical case.

    Examples
    --------
    >>> np.random.seed(123)
    >>> df = pd.DataFrame({"A1970" : {0 : "a", 1 : "b", 2 : "c"},
    ...                    "A1980" : {0 : "d", 1 : "e", 2 : "f"},
    ...                    "B1970" : {0 : 2.5, 1 : 1.2, 2 : .7},
    ...                    "B1980" : {0 : 3.2, 1 : 1.3, 2 : .1},
    ...                    "X"     : dict(zip(range(3), np.random.randn(3)))
    ...                   })
    >>> df["id"] = df.index
    >>> df
      A1970 A1980  B1970  B1980         X  id
    0     a     d    2.5    3.2 -1.085631   0
    1     b     e    1.2    1.3  0.997345   1
    2     c     f    0.7    0.1  0.282978   2
    >>> pd.wide_to_long(df, ["A", "B"], i="id", j="year")
    ... # doctest: +NORMALIZE_WHITESPACE
                    X  A    B
    id year
    0  1970 -1.085631  a  2.5
    1  1970  0.997345  b  1.2
    2  1970  0.282978  c  0.7
    0  1980 -1.085631  d  3.2
    1  1980  0.997345  e  1.3
    2  1980  0.282978  f  0.1

    With multiple id columns

    >>> df = pd.DataFrame({
    ...     'famid': [1, 1, 1, 2, 2, 2, 3, 3, 3],
    ...     'birth': [1, 2, 3, 1, 2, 3, 1, 2, 3],
    ...     'ht1': [2.8, 2.9, 2.2, 2, 1.8, 1.9, 2.2, 2.3, 2.1],
    ...     'ht2': [3.4, 3.8, 2.9, 3.2, 2.8, 2.4, 3.3, 3.4, 2.9]
    ... })
    >>> df
       birth  famid  ht1  ht2
    0      1      1  2.8  3.4
    1      2      1  2.9  3.8
    2      3      1  2.2  2.9
    3      1      2  2.0  3.2
    4      2      2  1.8  2.8
    5      3      2  1.9  2.4
    6      1      3  2.2  3.3
    7      2      3  2.3  3.4
    8      3      3  2.1  2.9
    >>> l = pd.wide_to_long(df, stubnames='ht', i=['famid', 'birth'], j='age')
    >>> l
    ... # doctest: +NORMALIZE_WHITESPACE
                      ht
    famid birth age
    1     1     1    2.8
                2    3.4
          2     1    2.9
                2    3.8
          3     1    2.2
                2    2.9
    2     1     1    2.0
                2    3.2
          2     1    1.8
                2    2.8
          3     1    1.9
                2    2.4
    3     1     1    2.2
                2    3.3
          2     1    2.3
                2    3.4
          3     1    2.1
                2    2.9

    Going from long back to wide just takes some creative use of `unstack`

    >>> w = l.unstack()
    >>> w.columns = w.columns.map('{0[0]}{0[1]}'.format)
    >>> w.reset_index()
       famid  birth  ht1  ht2
    0      1      1  2.8  3.4
    1      1      2  2.9  3.8
    2      1      3  2.2  2.9
    3      2      1  2.0  3.2
    4      2      2  1.8  2.8
    5      2      3  1.9  2.4
    6      3      1  2.2  3.3
    7      3      2  2.3  3.4
    8      3      3  2.1  2.9

    Less wieldy column names are also handled

    >>> np.random.seed(0)
    >>> df = pd.DataFrame({'A(quarterly)-2010': np.random.rand(3),
    ...                    'A(quarterly)-2011': np.random.rand(3),
    ...                    'B(quarterly)-2010': np.random.rand(3),
    ...                    'B(quarterly)-2011': np.random.rand(3),
    ...                    'X' : np.random.randint(3, size=3)})
    >>> df['id'] = df.index
    >>> df # doctest: +NORMALIZE_WHITESPACE, +ELLIPSIS
       A(quarterly)-2010  A(quarterly)-2011  B(quarterly)-2010  ...
    0           0.548814           0.544883           0.437587  ...
    1           0.715189           0.423655           0.891773  ...
    2           0.602763           0.645894           0.963663  ...
       X  id
    0  0   0
    1  1   1
    2  1   2

    >>> pd.wide_to_long(df, ['A(quarterly)', 'B(quarterly)'], i='id',
    ...                 j='year', sep='-')
    ... # doctest: +NORMALIZE_WHITESPACE
             X  A(quarterly)  B(quarterly)
    id year
    0  2010  0      0.548814     0.437587
    1  2010  1      0.715189     0.891773
    2  2010  1      0.602763     0.963663
    0  2011  0      0.544883     0.383442
    1  2011  1      0.423655     0.791725
    2  2011  1      0.645894     0.528895

    If we have many columns, we could also use a regex to find our
    stubnames and pass that list on to wide_to_long

    >>> stubnames = sorted(
    ...     set([match[0] for match in df.columns.str.findall(
    ...         r'[A-B]\(.*\)').values if match != [] ])
    ... )
    >>> list(stubnames)
    ['A(quarterly)', 'B(quarterly)']

    All of the above examples have integers as suffixes. It is possible to
    have non-integers as suffixes.

    >>> df = pd.DataFrame({
    ...     'famid': [1, 1, 1, 2, 2, 2, 3, 3, 3],
    ...     'birth': [1, 2, 3, 1, 2, 3, 1, 2, 3],
    ...     'ht_one': [2.8, 2.9, 2.2, 2, 1.8, 1.9, 2.2, 2.3, 2.1],
    ...     'ht_two': [3.4, 3.8, 2.9, 3.2, 2.8, 2.4, 3.3, 3.4, 2.9]
    ... })
    >>> df
       birth  famid  ht_one  ht_two
    0      1      1     2.8     3.4
    1      2      1     2.9     3.8
    2      3      1     2.2     2.9
    3      1      2     2.0     3.2
    4      2      2     1.8     2.8
    5      3      2     1.9     2.4
    6      1      3     2.2     3.3
    7      2      3     2.3     3.4
    8      3      3     2.1     2.9

    >>> l = pd.wide_to_long(df, stubnames='ht', i=['famid', 'birth'], j='age',
                            sep='_', suffix='\w')
    >>> l
    ... # doctest: +NORMALIZE_WHITESPACE
                      ht
    famid birth age
    1     1     one  2.8
                two  3.4
          2     one  2.9
                two  3.8
          3     one  2.2
                two  2.9
    2     1     one  2.0
                two  3.2
          2     one  1.8
                two  2.8
          3     one  1.9
                two  2.4
    3     1     one  2.2
                two  3.3
          2     one  2.3
                two  3.4
          3     one  2.1
                two  2.9

def wide_to_long(df, stubnames, i, j, sep="", suffix=r'\d+'):
    r"""
    Wide panel to long format. Less flexible but more user-friendly than melt.

    With stubnames ['A', 'B'], this function expects to find one or more
    group of columns with format
    A-suffix1, A-suffix2,..., B-suffix1, B-suffix2,...
    You specify what you want to call this suffix in the resulting long format
    with `j` (for example `j='year'`)

    Each row of these wide variables are assumed to be uniquely identified by
    `i` (can be a single column name or a list of column names)

    All remaining variables in the data frame are left intact.

    Parameters
    ----------
    df : DataFrame
        The wide-format DataFrame
    stubnames : str or list-like
        The stub name(s). The wide format variables are assumed to
        start with the stub names.
    i : str or list-like
        Column(s) to use as id variable(s)
    j : str
        The name of the sub-observation variable. What you wish to name your
        suffix in the long format.
    sep : str, default ""
        A character indicating the separation of the variable names
        in the wide format, to be stripped from the names in the long format.
        For example, if your column names are A-suffix1, A-suffix2, you
        can strip the hyphen by specifying `sep='-'`

        .. versionadded:: 0.20.0

    suffix : str, default '\\d+'
        A regular expression capturing the wanted suffixes. '\\d+' captures
        numeric suffixes. Suffixes with no numbers could be specified with the
        negated character class '\\D+'. You can also further disambiguate
        suffixes, for example, if your wide variables are of the form
        A-one, B-two,.., and you have an unrelated column A-rating, you can
        ignore the last one by specifying `suffix='(!?one|two)'`

        .. versionadded:: 0.20.0

        .. versionchanged:: 0.23.0
            When all suffixes are numeric, they are cast to int64/float64.

    Returns
    -------
    DataFrame
        A DataFrame that contains each stub name as a variable, with new index
        (i, j).

    Notes
    -----
    All extra variables are left untouched. This simply uses
    `pandas.melt` under the hood, but is hard-coded to "do the right thing"
    in a typical case.

    Examples
    --------
    >>> np.random.seed(123)
    >>> df = pd.DataFrame({"A1970" : {0 : "a", 1 : "b", 2 : "c"},
    ...                    "A1980" : {0 : "d", 1 : "e", 2 : "f"},
    ...                    "B1970" : {0 : 2.5, 1 : 1.2, 2 : .7},
    ...                    "B1980" : {0 : 3.2, 1 : 1.3, 2 : .1},
    ...                    "X"     : dict(zip(range(3), np.random.randn(3)))
    ...                   })
    >>> df["id"] = df.index
    >>> df
      A1970 A1980  B1970  B1980         X  id
    0     a     d    2.5    3.2 -1.085631   0
    1     b     e    1.2    1.3  0.997345   1
    2     c     f    0.7    0.1  0.282978   2
    >>> pd.wide_to_long(df, ["A", "B"], i="id", j="year")
    ... # doctest: +NORMALIZE_WHITESPACE
                    X  A    B
    id year
    0  1970 -1.085631  a  2.5
    1  1970  0.997345  b  1.2
    2  1970  0.282978  c  0.7
    0  1980 -1.085631  d  3.2
    1  1980  0.997345  e  1.3
    2  1980  0.282978  f  0.1

    With multiple id columns

    >>> df = pd.DataFrame({
    ...     'famid': [1, 1, 1, 2, 2, 2, 3, 3, 3],
    ...     'birth': [1, 2, 3, 1, 2, 3, 1, 2, 3],
    ...     'ht1': [2.8, 2.9, 2.2, 2, 1.8, 1.9, 2.2, 2.3, 2.1],
    ...     'ht2': [3.4, 3.8, 2.9, 3.2, 2.8, 2.4, 3.3, 3.4, 2.9]
    ... })
    >>> df
       birth  famid  ht1  ht2
    0      1      1  2.8  3.4
    1      2      1  2.9  3.8
    2      3      1  2.2  2.9
    3      1      2  2.0  3.2
    4      2      2  1.8  2.8
    5      3      2  1.9  2.4
    6      1      3  2.2  3.3
    7      2      3  2.3  3.4
    8      3      3  2.1  2.9
    >>> l = pd.wide_to_long(df, stubnames='ht', i=['famid', 'birth'], j='age')
    >>> l
    ... # doctest: +NORMALIZE_WHITESPACE
                      ht
    famid birth age
    1     1     1    2.8
                2    3.4
          2     1    2.9
                2    3.8
          3     1    2.2
                2    2.9
    2     1     1    2.0
                2    3.2
          2     1    1.8
                2    2.8
          3     1    1.9
                2    2.4
    3     1     1    2.2
                2    3.3
          2     1    2.3
                2    3.4
          3     1    2.1
                2    2.9

    Going from long back to wide just takes some creative use of `unstack`

    >>> w = l.unstack()
    >>> w.columns = w.columns.map('{0[0]}{0[1]}'.format)
    >>> w.reset_index()
       famid  birth  ht1  ht2
    0      1      1  2.8  3.4
    1      1      2  2.9  3.8
    2      1      3  2.2  2.9
    3      2      1  2.0  3.2
    4      2      2  1.8  2.8
    5      2      3  1.9  2.4
    6      3      1  2.2  3.3
    7      3      2  2.3  3.4
    8      3      3  2.1  2.9

    Less wieldy column names are also handled

    >>> np.random.seed(0)
    >>> df = pd.DataFrame({'A(quarterly)-2010': np.random.rand(3),
    ...                    'A(quarterly)-2011': np.random.rand(3),
    ...                    'B(quarterly)-2010': np.random.rand(3),
    ...                    'B(quarterly)-2011': np.random.rand(3),
    ...                    'X' : np.random.randint(3, size=3)})
    >>> df['id'] = df.index
    >>> df # doctest: +NORMALIZE_WHITESPACE, +ELLIPSIS
       A(quarterly)-2010  A(quarterly)-2011  B(quarterly)-2010  ...
    0           0.548814           0.544883           0.437587  ...
    1           0.715189           0.423655           0.891773  ...
    2           0.602763           0.645894           0.963663  ...
       X  id
    0  0   0
    1  1   1
    2  1   2

    >>> pd.wide_to_long(df, ['A(quarterly)', 'B(quarterly)'], i='id',
    ...                 j='year', sep='-')
    ... # doctest: +NORMALIZE_WHITESPACE
             X  A(quarterly)  B(quarterly)
    id year
    0  2010  0      0.548814     0.437587
    1  2010  1      0.715189     0.891773
    2  2010  1      0.602763     0.963663
    0  2011  0      0.544883     0.383442
    1  2011  1      0.423655     0.791725
    2  2011  1      0.645894     0.528895

    If we have many columns, we could also use a regex to find our
    stubnames and pass that list on to wide_to_long

    >>> stubnames = sorted(
    ...     set([match[0] for match in df.columns.str.findall(
    ...         r'[A-B]\(.*\)').values if match != [] ])
    ... )
    >>> list(stubnames)
    ['A(quarterly)', 'B(quarterly)']

    All of the above examples have integers as suffixes. It is possible to
    have non-integers as suffixes.

    >>> df = pd.DataFrame({
    ...     'famid': [1, 1, 1, 2, 2, 2, 3, 3, 3],
    ...     'birth': [1, 2, 3, 1, 2, 3, 1, 2, 3],
    ...     'ht_one': [2.8, 2.9, 2.2, 2, 1.8, 1.9, 2.2, 2.3, 2.1],
    ...     'ht_two': [3.4, 3.8, 2.9, 3.2, 2.8, 2.4, 3.3, 3.4, 2.9]
    ... })
    >>> df
       birth  famid  ht_one  ht_two
    0      1      1     2.8     3.4
    1      2      1     2.9     3.8
    2      3      1     2.2     2.9
    3      1      2     2.0     3.2
    4      2      2     1.8     2.8
    5      3      2     1.9     2.4
    6      1      3     2.2     3.3
    7      2      3     2.3     3.4
    8      3      3     2.1     2.9

    >>> l = pd.wide_to_long(df, stubnames='ht', i=['famid', 'birth'], j='age',
                            sep='_', suffix='\w')
    >>> l
    ... # doctest: +NORMALIZE_WHITESPACE
                      ht
    famid birth age
    1     1     one  2.8
                two  3.4
          2     one  2.9
                two  3.8
          3     one  2.2
                two  2.9
    2     1     one  2.0
                two  3.2
          2     one  1.8
                two  2.8
          3     one  1.9
                two  2.4
    3     1     one  2.2
                two  3.3
          2     one  2.3
                two  3.4
          3     one  2.1
                two  2.9
    """
    def get_var_names(df, stub, sep, suffix):
        regex = r'^{stub}{sep}{suffix}$'.format(
            stub=re.escape(stub), sep=re.escape(sep), suffix=suffix)
        pattern = re.compile(regex)
        return [col for col in df.columns if pattern.match(col)]

    def melt_stub(df, stub, i, j, value_vars, sep):
        newdf = melt(df, id_vars=i, value_vars=value_vars,
                     value_name=stub.rstrip(sep), var_name=j)
        newdf[j] = Categorical(newdf[j])
        newdf[j] = newdf[j].str.replace(re.escape(stub + sep), "")

        # GH17627 Cast numerics suffixes to int/float
        newdf[j] = to_numeric(newdf[j], errors='ignore')

        return newdf.set_index(i + [j])

    if not is_list_like(stubnames):
        stubnames = [stubnames]
    else:
        stubnames = list(stubnames)

    if any(col in stubnames for col in df.columns):
        raise ValueError("stubname can't be identical to a column name")

    if not is_list_like(i):
        i = [i]
    else:
        i = list(i)

    if df[i].duplicated().any():
        raise ValueError("the id variables need to uniquely identify each row")

    value_vars = [get_var_names(df, stub, sep, suffix) for stub in stubnames]

    value_vars_flattened = [e for sublist in value_vars for e in sublist]
    id_vars = list(set(df.columns.tolist()).difference(value_vars_flattened))

    melted = [melt_stub(df, s, i, j, v, sep)
              for s, v in zip(stubnames, value_vars)]
    melted = melted[0].join(melted[1:], how='outer')

    if len(i) == 1:
        new = df[id_vars].set_index(i).join(melted)
        return new

    new = df[id_vars].merge(melted.reset_index(), on=i).set_index(i + [j])

    return new