Statistics for an Athletic Association
Codewars Kata 13√
Description
Link: stat_CAA
-简述：本题以时|分|秒的数字形式给定一系列长跑运动员的时间成绩，求出其范围、平均值和中位数。
-思路：将数字形式转换为具体时间，求range average medium，然后再将他们转换为数字形式。
-难点：1 format中数字的表达形式
You are the “computer expert” of a local Athletic Association (C.A.A.). Many teams of runners come to compete. Each time you get a string of all race results of every team who has run. For example here is a string showing the individual results of a team of 5 runners:
| “01 | 15 | 59, 1 | 47 | 6, 01 | 17 | 20, 1 | 32 | 34, 2 | 3 | 17” |
| Each part of the string is of the form: h | m | s where h, m, s (h for hour, m for minutes, s for seconds) are positive or null integer (represented as strings) with one or two digits. There are no traps in this format. |
To compare the results of the teams you are asked for giving three statistics; range, average and median.
Range : difference between the lowest and highest values. In {4, 6, 9, 3, 7} the lowest value is 3, and the highest is 9, so the range is 9 − 3 = 6.
Mean or Average : To calculate mean, add together all of the numbers in a set and then divide the sum by the total count of numbers.
Median : In statistics, the median is the number separating the higher half of a data sample from the lower half. The median of a finite list of numbers can be found by arranging all the observations from lowest value to highest value and picking the middle one (e.g., the median of {3, 3, 5, 9, 11} is 5) when there is an odd number of observations. If there is an even number of observations, then there is no single middle value; the median is then defined to be the mean of the two middle values (the median of {3, 5, 6, 9} is (5 + 6) / 2 = 5.5).
Your task is to return a string giving these 3 values. For the example given above, the string result will be
“Range: 00|47|18 Average: 01|35|15 Median: 01|32|34”of the form: “Range: hh|mm|ss Average: hh|mm|ss Median: hh|mm|ss” where hh, mm, ss are integers (represented by strings) with each 2 digits.
Remarks: if a result in seconds is ab.xy… it will be given truncated as ab. if the given string is “” you will return “”
My solution
def changetoHMS(x):
h1 = int(x // 3600)
m1 = int((x - 3600 * h1) // 60)
s1 = int(x - 3600 * h1 - 60 * m1)
return ' {h1:02d}|{m1:02d}|{s1:02d}'.format(**locals())
def changetoTime(str):
b1 = str.split('|')
return int(b1[0]) * 3600 + int(b1[1]) * 60 + int(b1[2])
def getMedium(lst):
l = len(lst)
lst.sort()
if l % 2 == 0:
return (lst[l // 2] + lst[l // 2 - 1]) / 2
else:
return lst[l // 2]
def stat(strg):
if strg =='':
return ''
else:
s = strg.split(',')
lst = []
for i in range(0,len(s)):
lst.append(changetoTime(s[i]))
avgr = changetoHMS(sum(lst)//len(lst))
rger = changetoHMS(max(lst)-min(lst))
medr = changetoHMS(getMedium(lst))
return 'Range:{rger} Average:{avgr} Median:{medr}'.format(**locals())
print(stat("01|15|59, 1|47|16, 01|17|20, 1|32|34, 2|17|17"))
Given solutions
def stat(strg):
def get_time(s):
'''Returns the time, in seconds, represented by s.'''
hh, mm, ss = [int(v) for v in s.split('|')]
return hh * 3600 + mm * 60 + ss
def format_time(time):
'''Returns the given time as a string in the form "hh|mm|ss".'''
hh = time // 3600
mm = time // 60 % 60
ss = time % 60
return '{hh:02d}|{mm:02d}|{ss:02d}'.format(**locals())
def get_range(times):
return times[-1] - times[0]
def get_average(times):
return sum(times) // len(times)
def get_median(times):
middle = len(times) >> 1
return (times[middle] if len(times) & 1 else
(times[middle - 1] + times[middle]) // 2)
if strg == '':
return strg
times = [get_time(s) for s in strg.split(', ')]
times.sort()
rng = format_time(get_range(times))
avg = format_time(get_average(times))
mdn = format_time(get_median(times))
return 'Range: {rng} Average: {avg} Median: {mdn}'.format(**locals())
Points
1 报错 ValueError: Unknown format code ‘d’ for object of type ‘float’ 加入format中的参数不是整数，float无法用d表示。
2 format()函数讲解：format()
3 运行时出现以下错误：TypeError: ‘int’ object is not callable 报错处前面定义了一个与函数同名的变量，后期再遇到该函数时，则会报错。将变量改名即可。
