pyTony 888 Sometimes we want to tackle some problem specifying properties of sequence of numbers consisting the integer. Then it is usefull to have function to join many integers to make a long integer. Most efficient way for tight loops is to stay in integers domain and not go through the string representation. Inspired by C++ question and my answer to it: http://www.daniweb.com/software-development/cpp/threads/425472/how-concatenate-number-in-c#post1819908
Edited by pyTony
def together(*numbers):
b = numbers[-1]
for a in reversed(numbers[:-1]):
magnitude = 1
while b >= magnitude:
magnitude *= 10
b = magnitude * a + b
return b
print together(123, 456, 100, 789, 10, 999, 1000, 99)Code snippet updated to handle also zero values:
def together(*numbers):
b = numbers[-1]
previous_zeroes = 1
for a in reversed(numbers[:-1]):
if not a:
previous_zeroes *= 10
else:
magnitude = 10
while b >= magnitude:
magnitude *= 10
b = previous_zeroes * magnitude * a + b
previous_zeroes = 1
return b
if __name__ == '__main__':
print together(123, 456, 0, 0, 789, 10, 123, 1000, 99, 0)
Lucaci Andrew 140 you could also do it like this
def nr(*nrs):
nr = ""
for i in range (0, len(nrs)):
a = str(nrs[i])
nr = nr + a
return int(nr)
print nr(1, 2, 3, 4, 5, 6) + 100000000
since Python works with strings alot.
and if you want to put a 0 in front, the program should ignore it.
def nr(*nrs):
nr = ""
for i in range (0, len(nrs)):
if nrs[i] == '0': continue
else :
a = str(nrs[i])
nr = nr + a
return int(nr)
print nr(0, 2, 3, 4, 5, 0, 6)
thow, the int(nr) will do that, I put this to evidentiate if so the need.
Edited by Lucaci Andrew
pyTony 888 As I mentioned, keeping in numbers is likely to be most efficient, but doing with strings is super simple:
def nr(*nrs):
return int(''.join(str(n) for n in nrs))
print(nr(1, 2, 3, 4, 5, 6) + 100000000)
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