python 自然语言分析第六周作业

"""写一个程序按长度对词排序。定义一个辅助函数 cmp_len，它在词长上使用 cmp
比较函数"""
def cmp_len(a,b):
if len(a)>=len(b):
s=a
if len(a)<len(b):
s=b
return s
words=['I','love','NPL']
def cmp(words):
words.sort(key=len)
return words
cmp(words)
"""一个函数处理一个词链表（含重复项），返回一个按照频率递减排序的词链表（没
有重复项）。例如：如果输入链表中包含词table的 10个实例， chair的 9个实例， 那
么在输出链表中table会出现在chair前面"""
word=["1","2","3","1","2","3","2","3","3"]
def get_feq(word):
dic={}
for i in range(len(word)):
dic[word[i]]=word.count(word[i])

return sorted(dic.keys(),reverse=True)
get_feq(word)

"""◑从 Python的标准库的operator模块导入itemgetter()函数（ 即from operator i
mport itemgetter）。 创建一个包含几个词的链表words。现在尝试调用： sorted(w
ords, key=itemgetter(1))和sorted(words, key=itemgetter(-1))。解释item
getter()正在做什么"""
from operator import itemgetter
words=["I","LOVE","YOU"]
sorted(words, key=itemgetter(1))
sorted(words, key=itemgetter(-1))
#获取对象的第几个值
"""写一个程序实现发现一种n×n的四方联词的蛮力算法：纵横字谜，它的第n行的词
与第 n列的词相同。 讨论见http://itre.cis.upenn.edu/ myl/languagelog/archives/002679.ht
ml"""
s="The moment you think about giving up,think of the reason why you held on so long"
word=["The","moment","you","think","about","giving","up"]

def zonghengzimi(word):
#生成以word为基础的，10X10的10阶乘
import pandas as pd
import numpy as np
import random
raw=[]
for i in range(100):
raw.append(chr(random.randint(97, 122)))
data=pd.DataFrame(np.array(raw).reshape(10,10))
print (len(data))
print (data  )
#生成了10X10的阶乘随机字母，需要把七个词语放进去，并把一部分改成空白元素进行填写
for i in range(len(word)):
a=random.randint(1,10)
b=random.randint(1,10)
for j in range(len(word[i])):

if a>b and b+len(word[i])<10:
data.ix[a,b+j]=word[i][j]
if a>b and b+len(word[i])>10:
b=0
data.ix[a,b+j]=word[i][j]
return data
#这样就简单生成了一个纵横字谜,尝试在这些子母中找出相应的word当中的单词
zonghengzimi(word)