如果它们出现与斐波那契序列相同的次数,我试图只打印出单词。如果一个单词显示1,2,3,5,8等,那么它将打印出来。我已经让程序根据出现的次数打印出单词。我无法弄清楚如何在程序中使用序列。任何提示或示例都将非常感激。
def fib():
a,b = 0, 1
while 1:
yield a
a, b= b, a+b
from collections import Counter
import string
while True:
filename=raw_input('Enter a file name: ')
if filename == 'exit':
break
try:
file = open(filename, 'r')
text=file.read()
file.close()
except:
print('file does not exist')
else:
for word in string.punctuation:
text=text.replace(word, "")
word_list = text.lower().split(None)
word_freq = {}
for word in word_list:
if len(word) > 1:
word_freq[word] = word_freq.get(word, 0) + 1
print(sorted(word_freq.items(), key=lambda item: item[1]))
// I am pretty sure something with the seqeunce should go into the above line
// but have been unable to figure it out.
print('Bye')
答案 0 :(得分:3)
class FibSet:
'''Fibonacci sequence with the in operator defined'''
def __init__(self):
self.a, self.b = 0, 1
self.fib = set()
def __contains__(self, n):
if n > self.b:
self.compute_upto(n)
return n in self.fib
def compute_upto(self, n):
while self.b < n:
self.fib.add(self.a)
self.a, self.b = self.b, self.a + self.b
答案 1 :(得分:0)
你可以将其解决到最后:
frequencies = sorted(word_freq.items(), key=lambda item: item[1])
def fib2(n):
phi = (1.0 + sqrt(5)) / 2.0
return (phi**n - (1 - phi)**n) / (sqrt(5))
for word in frequencies:
n = 0
while word[1] < fib2(n):
n += 1
if word[1] == fib2(n):
print word[0]
您必须事先import * from math
,因为我正在使用nth
斐波纳契数的闭合函数。
答案 2 :(得分:0)
这是一种可以使用的方法,可以最大限度地减少对现有代码的影响。
import itertools
...
items = [item for item in word_freq.items() if
item[1] in itertools.takewhile(lambda f: f <= item[1], fib())]
print(sorted(items, key=lambda item: item[1]))
但是在阅读文件之前制作一组斐波那契数字可能更有意义。