pylib/cqlshlib/util.py

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#
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from itertools import izip

def split_list(items, pred):
    """
    Split up a list (or other iterable) on the elements which satisfy the
    given predicate 'pred'. Elements for which 'pred' returns true start a new
    sublist for subsequent elements, which will accumulate in the new sublist
    until the next satisfying element.

    >>> split_list([0, 1, 2, 5, 99, 8], lambda n: (n % 2) == 0)
    [[0], [1, 2], [5, 99, 8], []]
    """

    thisresult = []
    results = [thisresult]
    for i in items:
        thisresult.append(i)
        if pred(i):
            thisresult = []
            results.append(thisresult)
    return results

def find_common_prefix(strs):
    """
    Given a list (iterable) of strings, return the longest common prefix.

    >>> find_common_prefix(['abracadabra', 'abracadero', 'abranch'])
    'abra'
    >>> find_common_prefix(['abracadabra', 'abracadero', 'mt. fuji'])
    ''
    """

    common = []
    for cgroup in izip(*strs):
        if all(x == cgroup[0] for x in cgroup[1:]):
            common.append(cgroup[0])
        else:
            break
    return ''.join(common)

def list_bifilter(pred, iterable):
    """
    Filter an iterable into two output lists: the first containing all
    elements of the iterable for which 'pred' returns true, and the second
    containing all others. Order of the elements is otherwise retained.

    >>> list_bifilter(lambda x: isinstance(x, int), (4, 'bingo', 1.2, 6, True))
    ([4, 6], ['bingo', 1.2, True])
    """

    yes_s = []
    no_s = []
    for i in iterable:
        (yes_s if pred(i) else no_s).append(i)
    return yes_s, no_s

def identity(x):
    return x

def trim_if_present(s, prefix):
    if s.startswith(prefix):
        return s[len(prefix):]
    return s