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
Joining integer numbers together to make new number
About the Author
TrustyTony
888
pyMod
IT Pro doing Eng-Fin-Eng translations
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)
TrustyTony
888
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.
TrustyTony
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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