How to enumerate a range of numbers starting at 1
Question:
I am using Python 2.5, I want an enumeration like so (starting at 1 instead of 0):
[(1, 2000), (2, 2001), (3, 2002), (4, 2003), (5, 2004)]
I know in Python 2.6 you can do: h = enumerate(range(2000, 2005), 1) to give the above result but in python2.5 you cannot…
Using Python 2.5:
>>> h = enumerate(range(2000, 2005))
>>> [x for x in h]
[(0, 2000), (1, 2001), (2, 2002), (3, 2003), (4, 2004)]
Does anyone know a way to get that desired result in Python 2.5?
Answers:
h = [(i + 1, x) for i, x in enumerate(xrange(2000, 2005))]
As you already mentioned, this is straightforward to do in Python 2.6 or newer:
enumerate(range(2000, 2005), 1)
Python 2.5 and older do not support the start parameter so instead you could create two range objects and zip them:
r = xrange(2000, 2005)
r2 = xrange(1, len(r) + 1)
h = zip(r2, r)
print h
Result:
[(1, 2000), (2, 2001), (3, 2002), (4, 2003), (5, 2004)]
If you want to create a generator instead of a list then you can use izip instead.
>>> h = enumerate(range(2000, 2005))
>>> [(tup[0]+1, tup[1]) for tup in h]
[(1, 2000), (2, 2001), (3, 2002), (4, 2003), (5, 2004)]
Since this is somewhat verbose, I’d recommend writing your own function to generalize it:
def enumerate_at(xs, start):
return ((tup[0]+start, tup[1]) for tup in enumerate(xs))
from itertools import count, izip
def enumerate(L, n=0):
return izip( count(n), L)
# if 2.5 has no count
def count(n=0):
while True:
yield n
n+=1
Now h = list(enumerate(xrange(2000, 2005), 1)) works.
Easy, just define your own function that does what you want:
def enum(seq, start=0):
for i, x in enumerate(seq):
yield i+start, x
enumerate is trivial, and so is re-implementing it to accept a start:
def enumerate(iterable, start = 0):
n = start
for i in iterable:
yield n, i
n += 1
Note that this doesn’t break code using enumerate without start argument. Alternatively, this oneliner may be more elegant and possibly faster, but breaks other uses of enumerate:
enumerate = ((index+1, item) for index, item)
The latter was pure nonsense. @Duncan got the wrapper right.
>>> list(enumerate(range(1999, 2005)))[1:]
[(1, 2000), (2, 2001), (3, 2002), (4, 2003), (5, 2004)]
Simplest way to do in Python 2.5 exactly what you ask about:
import itertools as it
... it.izip(it.count(1), xrange(2000, 2005)) ...
If you want a list, as you appear to, use zip in lieu of it.izip.
(BTW, as a general rule, the best way to make a list out of a generator or any other iterable X is not [x for x in X], but rather list(X)).
Ok, I feel a bit stupid here… what’s the reason not to just do it with something like
[(a+1,b) for (a,b) in enumerate(r)] ? If you won’t function, no problem either:
>>> r = range(2000, 2005)
>>> [(a+1,b) for (a,b) in enumerate(r)]
[(1, 2000), (2, 2001), (3, 2002), (4, 2003), (5, 2004)]
>>> enumerate1 = lambda r:((a+1,b) for (a,b) in enumerate(r))
>>> list(enumerate1(range(2000,2005))) # note - generator just like original enumerate()
[(1, 2000), (2, 2001), (3, 2002), (4, 2003), (5, 2004)]
Just to put this here for posterity sake, in 2.6 the “start” parameter was added to enumerate like so:
enumerate(sequence, start=1)
I don’t know how these posts could possibly be made more complicated then the following:
# Just pass the start argument to enumerate ...
for i,word in enumerate(allWords, 1):
word2idx[word]=i
idx2word[i]=word
Python 3
Official Python documentation: enumerate(iterable, start=0)
You don’t need to write your own generator as other answers here suggest. The built-in Python standard library already contains a function that does exactly what you want:
>>> seasons = ['Spring', 'Summer', 'Fall', 'Winter']
>>> list(enumerate(seasons))
[(0, 'Spring'), (1, 'Summer'), (2, 'Fall'), (3, 'Winter')]
>>> list(enumerate(seasons, start=1))
[(1, 'Spring'), (2, 'Summer'), (3, 'Fall'), (4, 'Winter')]
The built-in function is equivalent to this:
def enumerate(sequence, start=0):
n = start
for elem in sequence:
yield n, elem
n += 1
I am using Python 2.5, I want an enumeration like so (starting at 1 instead of 0):
[(1, 2000), (2, 2001), (3, 2002), (4, 2003), (5, 2004)]
I know in Python 2.6 you can do: h = enumerate(range(2000, 2005), 1) to give the above result but in python2.5 you cannot…
Using Python 2.5:
>>> h = enumerate(range(2000, 2005))
>>> [x for x in h]
[(0, 2000), (1, 2001), (2, 2002), (3, 2003), (4, 2004)]
Does anyone know a way to get that desired result in Python 2.5?
h = [(i + 1, x) for i, x in enumerate(xrange(2000, 2005))]
As you already mentioned, this is straightforward to do in Python 2.6 or newer:
enumerate(range(2000, 2005), 1)
Python 2.5 and older do not support the start parameter so instead you could create two range objects and zip them:
r = xrange(2000, 2005)
r2 = xrange(1, len(r) + 1)
h = zip(r2, r)
print h
Result:
[(1, 2000), (2, 2001), (3, 2002), (4, 2003), (5, 2004)]
If you want to create a generator instead of a list then you can use izip instead.
>>> h = enumerate(range(2000, 2005))
>>> [(tup[0]+1, tup[1]) for tup in h]
[(1, 2000), (2, 2001), (3, 2002), (4, 2003), (5, 2004)]
Since this is somewhat verbose, I’d recommend writing your own function to generalize it:
def enumerate_at(xs, start):
return ((tup[0]+start, tup[1]) for tup in enumerate(xs))
from itertools import count, izip
def enumerate(L, n=0):
return izip( count(n), L)
# if 2.5 has no count
def count(n=0):
while True:
yield n
n+=1
Now h = list(enumerate(xrange(2000, 2005), 1)) works.
Easy, just define your own function that does what you want:
def enum(seq, start=0):
for i, x in enumerate(seq):
yield i+start, x
enumerate is trivial, and so is re-implementing it to accept a start:
def enumerate(iterable, start = 0):
n = start
for i in iterable:
yield n, i
n += 1
Note that this doesn’t break code using enumerate without start argument. Alternatively, this oneliner may be more elegant and possibly faster, but breaks other uses of enumerate:
enumerate = ((index+1, item) for index, item)
The latter was pure nonsense. @Duncan got the wrapper right.
>>> list(enumerate(range(1999, 2005)))[1:]
[(1, 2000), (2, 2001), (3, 2002), (4, 2003), (5, 2004)]
Simplest way to do in Python 2.5 exactly what you ask about:
import itertools as it
... it.izip(it.count(1), xrange(2000, 2005)) ...
If you want a list, as you appear to, use zip in lieu of it.izip.
(BTW, as a general rule, the best way to make a list out of a generator or any other iterable X is not [x for x in X], but rather list(X)).
Ok, I feel a bit stupid here… what’s the reason not to just do it with something like
[(a+1,b) for (a,b) in enumerate(r)] ? If you won’t function, no problem either:
>>> r = range(2000, 2005)
>>> [(a+1,b) for (a,b) in enumerate(r)]
[(1, 2000), (2, 2001), (3, 2002), (4, 2003), (5, 2004)]
>>> enumerate1 = lambda r:((a+1,b) for (a,b) in enumerate(r))
>>> list(enumerate1(range(2000,2005))) # note - generator just like original enumerate()
[(1, 2000), (2, 2001), (3, 2002), (4, 2003), (5, 2004)]
Just to put this here for posterity sake, in 2.6 the “start” parameter was added to enumerate like so:
enumerate(sequence, start=1)
I don’t know how these posts could possibly be made more complicated then the following:
# Just pass the start argument to enumerate ...
for i,word in enumerate(allWords, 1):
word2idx[word]=i
idx2word[i]=word
Python 3
Official Python documentation: enumerate(iterable, start=0)
You don’t need to write your own generator as other answers here suggest. The built-in Python standard library already contains a function that does exactly what you want:
>>> seasons = ['Spring', 'Summer', 'Fall', 'Winter']
>>> list(enumerate(seasons))
[(0, 'Spring'), (1, 'Summer'), (2, 'Fall'), (3, 'Winter')]
>>> list(enumerate(seasons, start=1))
[(1, 'Spring'), (2, 'Summer'), (3, 'Fall'), (4, 'Winter')]
The built-in function is equivalent to this:
def enumerate(sequence, start=0):
n = start
for elem in sequence:
yield n, elem
n += 1