Separating list of years into decades

Question

I have the following array:

array([1963, 1964, 1965, 1966, 1967, 1968, 1969, 1970, 1971,
       1972, 1973, 1974, 1975, 1976, 1977, 1978, 1979, 1980, 1981, 1982,
       1983, 1984, 1985, 1986, 1987, 1988, 1989, 1990, 1991, 1992, 1993,
       1994, 1995, 1996, 1997, 1998, 1999, 2000, 2001, 2002, 2003, 2004,
       2005, 2006, 2007, 2008, 2009, 2010, 2011, 2012, 2013], dtype=object)

I would like to divide it into separate lists, one for each decade (e.g. 1970-1979 is one decade).

Right now, I am looping through the years and dividing into separate lists. Is there a more pythonic way to go about it?

`itertools.groupby(array, key= lambda x : str(x)[:3])`? – NightShadeQueen Jul 17 '15 at 17:38 — NightShadeQueen, Jul 17 '15 at 17:38
Is this a `numpy` array? – TigerhawkT3 Jul 17 '15 at 17:39 — TigerhawkT3, Jul 17 '15 at 17:39

Cory Kramer · Accepted Answer · 2015-07-17T17:41:11.490

8

You can use itertools.groupby and divide the year by 10. This will essentially group by the decade. Then use that in a list comprehension and create a new array.

>>> import numpy as np
>>> from itertools import groupby
>>> np.array([list(g) for k,g in groupby(a, lambda i: i // 10)])
array([[1963, 1964, 1965, 1966, 1967, 1968, 1969],
       [1970, 1971, 1972, 1973, 1974, 1975, 1976, 1977, 1978, 1979],
       [1980, 1981, 1982, 1983, 1984, 1985, 1986, 1987, 1988, 1989],
       [1990, 1991, 1992, 1993, 1994, 1995, 1996, 1997, 1998, 1999],
       [2000, 2001, 2002, 2003, 2004, 2005, 2006, 2007, 2008, 2009],
       [2010, 2011, 2012, 2013]], dtype=object)

Note that groupby requires that your sequence is sorted (which it looks like your data is).

edited Jul 17 '15 at 17:41

answered Jul 17 '15 at 17:39

Cory Kramer

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great answer although probably worth noting that the original list **must** be sorted (lol either you are a very fast typer or you were already doing that :P) – Joran Beasley Jul 17 '15 at 17:40
Is there any way to write a sort function in that line? – nonamorando Jul 17 '15 at 17:42
can't he write instead of a, sorted(a)? – omri_saadon Jul 17 '15 at 17:43
@omri_saadon Yes exactly – Cory Kramer Jul 17 '15 at 17:43
btw, what is the difference between i // 10 and i / 10 ? – omri_saadon Jul 17 '15 at 17:47
1

@omri_saadon Depends on the Python version, but in Python 3.x `int / int` will still do *float division* but I want to force *integer division* so I used `//`. https://stackoverflow.com/questions/1535596/what-is-the-reason-for-having-in-python – Cory Kramer Jul 17 '15 at 17:49

score 7 · Answer 2 · answered Oct 06 '18 at 19:35

There is another simpler method and a trick with np.floor

    import numpy as np
    years = np.array([1963, 1964, 1965, 1966, 1967, 1968, 1969, 1970, 1971,
          1972, 1973, 1974, 1975, 1976, 1977, 1978, 1979, 1980, 1981, 1982,
          1983, 1984, 1985, 1986, 1987, 1988, 1989, 1990, 1991, 1992, 1993,
          1994, 1995, 1996, 1997, 1998, 1999, 2000, 2001, 2002, 2003, 2004,
          2005, 2006, 2007, 2008, 2009, 2010, 2011, 2012, 2013])

    decades = []
    for each in years:
        decade = int(np.floor(each / 10) * 10)
        decades.append(decade)

    print(set(decades))

score 0 · Answer 3 · edited Aug 24 '22 at 12:40

0

For a pandas data frame, you can use list comprehension for shorter code:

df["decade"]= [ int(np.floor(year/10) * 10) for year in np.array(df["year"])]

edited Aug 24 '22 at 12:40

Adrian Mole

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answered Aug 18 '22 at 10:08

Abdulazeez Saliu

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Separating list of years into decades

3 Answers3