1. Most popular name
As in the fifth lecture, we use the data of the National Data on the relative frequency of given names in the population of U.S. births, stored in a subdirectory named names of the working directory, in files named yobxxxx.txt with xxxx (the year of birth) ranging from 1880 to 2013. Write a program most_popular_name.py that prompts the user for a first name, and finds out the first year when this name was most popular in terms of frequency of names being given, as a female name and as a male name.
本題分析Eric的解法姿染，他的單次運行效率較高叠必，程序相對簡單谢谦。相比之下醉蚁，我自己的寫法單次效率低，適合做大量查詢员舵。主要思路區(qū)別是我的寫法將所有文件遍歷后輸出一個龐大的字典棒仍，包括了年份、性別抡蛙、名字护昧、計數(shù)、頻率信息溜畅，程序復(fù)雜捏卓，單次運行效率低，勝在字典建立后多次查詢速度快慈格。
import os
first_name = input('Enter a first name: ')
min_male_frequency, min_female_frequency = 0, 0
male_first_year, female_first_year = None, None
tallies = {'F': {}, 'M': {}}
records = {'F': {}, 'M': {}}
#字典的嵌套怠晴，第一個字典實現(xiàn)性別分類，第二個字典分別實現(xiàn)題目要求
for filename in os.listdir('names'):
if not filename.endswith('.txt'):
continue
year = int(filename[3:7])
with open('names' + '/' + filename) as name:
for gender in {'F', 'M'}:
tallies[gender][year] = 0
for line in name:
name, gender, count = line.split(',')
count = int(count)
tallies[gender][year] += count
if name == first_name:
records[gender][year] = count
#遍歷了所有文件浴捆，獲得了所有年份的總計數(shù)tallies蒜田，{'F': {1880: XXX, 1881:XXX...}, 'M': {1880:XXX...}}
#獲得了所有年份要查詢名字的records，{'F': {1880: XXX, 1881:XXX...}, 'M': {1880:XXX...}}
frequencies = dict.fromkeys(('M', 'F'))
#以M和F作為key选泻，創(chuàng)建新的字典冲粤，{'F': None, 'M': None}
for gender in {'F','M'}:
frequencies[gender] = [(records[gender][year] * 100 / tallies[gender][year], year) for year in records[gender]]
frequencies[gender].sort(reverse = True)
#生成frequencies的dict美莫，方法是逐年查詢records和tallies字典，例如1880年梯捕，records中查到那一年該名字的計數(shù)records[gender][year]
#tallies中查到那一年所有名字的總計數(shù)厢呵，相除取百分比得到frequencies的數(shù)據(jù)。并由大到小排序傀顾。
#{'F':[(XX.XX, 1880), (XX.XX, 1881)...], 'M':[(XX.XX, 1880), (XX.XX, 1881)...]}
if frequencies['F']:
min_female_frequency, female_first_year = frequencies['F'][0][0], frequencies['F'][0][1]
if frequencies['M']:
min_male_frequency, male_first_year = frequencies['M'][0][0], frequencies['M'][0][1]
if not female_first_year:
print(f'In all years, {first_name} was never given as a female name.')
else:
print(f'In terms of frequency, {first_name} was the most popular as a female name '
f'first in the year {female_first_year}.\n'
f'It then accounted for {min_female_frequency:.2f}% of all female names'
)
if not male_first_year:
print(f'In all years, {first_name} was never given as a male name.')
else:
print(f'In terms of frequency, {first_name} was the most popular as a male name'
f'first in the year {male_first_year}.\n'
f'It then accounted for {min_male_frequency:.2f}% of all male names'
)
2. The 9 puzzle
Dispatch the integers from 0 to 8, with 0 possibly changed to None, as a list of 3 lists of size 3, to represent a 9 puzzle. For instance, let
[[4, 0, 8], [1, 3, 7], [5, 2, 6]]
OR
[[4, None ,8], [1, 3, 7], [5, 2, 6]]
represent the 9 puzzle
4 8
1 3 7
5 2 6
with the 8 integers being printed on 8 tiles that are placed in a frame with one location being tile free. The aim is to slide tiles horizontally or vertically so as to eventually reach the configuration
1 2 3
4 5 6
7 8
It can be shown that the puzzle is solvable iff the permutation of the integers 1, . . . , 8, determined by reading those integers off the puzzle from top to bottom and from left to right, is even. This is clearly a necessary condition since:
--sliding a tile horizontally does not change the number of inversions;
--sliding a tile vertically changes the number of inversions by -2, 0 or 2;
--the parity of the identity is even.
Write a program nine_puzzle.py with a function:
--validate_9_puzzle(grid) that prints out whether or not grid is a valid representation of a solvable 9 puzzle
def test(grid):
if len(grid) != 3:
return
test = []
for row in grid:
for column in row:
test.append(column)
#把所有g(shù)rid里面的element放在一個list中
if None in test:
test[test.index(None)] = 0
#如果有None例书，改成0
if sorted(test) != list(range(9)):
return
#如果不是1-8的數(shù)字和0/None構(gòu)成拣度，則不成立潮罪。這里不要用.sort()谅猾，那樣會改變原來的grid順序
permutation = 0
for i in range(8):
for j in range(i + 1, 9):
if test[i] and test[j] and test[i] > test[j]:
permutation += 1
#判斷從前向后所有非零數(shù)字（0在題中代表可以和任何相鄰數(shù)字互換位置），如果前面的數(shù)字比后面的大嫉拐，則記錄加1
if permutation % 2:
return
else:
return grid
def validate_9_puzzle(grid):
if test(grid):
print('This is a valid 9 puzzle, and it is solvable')
else:
print('This is an invalid or unsolvable 9 puzzle')
