Printer error
Codewars Kata 4√
Description
Link: printer_error
-简述：本题假设一个打印机的错误检查，若打印内容不在字母a-m内，则报错，输出错误次数。
-思路：运用正则找出所有异常数字
-难点：无
In a factory a printer prints labels for boxes. For one kind of boxes the printer has to use colors which, for the sake of simplicity, are named with letters from a to m. The colors used by the printer are recorded in a control string. For example a “good” control string would be aaabbbbhaijjjm meaning that the printer used three times color a, four times color b, one time color h then one time color a… Sometimes there are problems: lack of colors, technical malfunction and a “bad” control string is produced e.g. aaaxbbbbyyhwawiwjjjwwm. You have to write a function printer_error which given a string will output the error rate of the printer as a string representing a rational whose numerator is the number of errors and the denominator the length of the control string. Don’t reduce this fraction to a simpler expression. The string has a length greater or equal to one and contains only letters from ato z.
Examples:
s=”aaabbbbhaijjjm” error_printer(s) => “0/14”
s=”aaaxbbbbyyhwawiwjjjwwm” error_printer(s) => “8/22”
My solution:
import re
def printer_error(s):
# your code
m = str(len(re.findall(r'[n-z]',s)))
q = str(len(s))# 总长度 要加str!
return m+'/‘+q # str字符串可以用+连接
s="aammmxyzz"
x = printer_error(s)
print(x)
Given solutions:
from re import sub
def printer_error(s):
return "{}/{}".format(len(sub("[a-m]",'',s)),len(s)) ---
def printer_error(s):
errors = 0
count = len(s)
for i in s:
if i > "m":
errors += 1
return str(errors) + "/" + str(count)
Points
1 可以直接用if i > “m”:表达m之后的字母
