python - Convert List of strings to ints without removing 0 -

so have list of strings want convert list of ints

['030', '031', '031', '031', '030', '031', '031', '032', '031', '032']

how should go new list not remove zeroes

i want this:

[030, 031, 031, 031, 030, 031, 031, 032, 031, 032] 

not this:

[30, 31, 31, 31, 30, 31, 31, 32, 31, 32] 

thanks

the value of int("030") same value of int("30"). leading 0 doesn't have semantic value when talking number - if want keep leading 0, no longer storing number, rather representation of number, needs string.

the solution, if need use in both ways, store in commonly used form, , convert need it. if need leading 0 (that is, need string), keep is, , call int() on values required.

if opposite true, can use string formatting number padded required number of digits:

>>> "{0:03d}".format(30) '030' 

if 0 padding not consistent (that is, it's impossible recover formatting int), might best keep in both forms:

>>> [(value, int(value)) value in values] [('030', 30), ('031', 31), ('031', 31), ('031', 31), ('030', 30), ('031', 31), ('031', 31), ('032', 32), ('031', 31), ('032', 32)] 

what method best depends entirely on situation.

edit: specific case, in comments, want this:

>>> current = ("233", "199", "016") >>> modifier = "031" >>> ["".join(part) part in zip(*list(zip(*current))[:-1] + [modifier])] ['230', '193', '011'] 

to break down, want replace third digit of each number relevant number in modifier. done here using zip() make columns of numbers - list(zip(*current)) gives [('2', '1', '0'), ('3', '9', '1'), ('3', '9', '6')] - replace last column modified one, , use zip() again give rows again. join individual digits strings.

note in 2.x, zip() gives list, don't need wrap call in list(). in 3.x, generator.