从保存在2D列表中的三元组单词中构造文本

Question

我当前有一个文本，其单词保存为2D列表中的三元组。

我的2D列表：

[['Python', 'is', 'an'], ['interpreted,', 'high-level', 'and'], ['general-purpose', 'programming', 'language.'], ["Python's", 'design', 'philosophy'], ['emphasizes', 'code', 'readability'], ['with', 'its', 'notable'], ['use', 'of', 'significant'], ['whitespace.', 'Its', 'language'], ['constructs', 'and', 'object-oriented'], ['approach', 'aim', 'to'], ['help', 'programmers', 'write'], ['clear,', 'logical', 'code'], ['for', 'small', 'and'], ['large-scale', 'projects.']]

我正在创建一个Python代码，它从这些三元组中随机选择一组，然后尝试使用最后两个单词并选择以这两个单词开头的三元组来创建一个新的随机文本。最后，当200个单词正在编写或无法选择其他三元组集时，我的程序结束。

到目前为止的代码：

import random
with open(r'c:\python4_TRIPLETSSample.txt', 'r') as file:
data = file.read().replace('n', '').split()
lines = [data[i:i + 3] for i in range(0, len(data), 3)]
random.shuffle([random.shuffle(i) for i in lines])
first_triplet = random.choice(lines)
last_two = first_triplet[1:3]

output_text=[]
while True:
candidates = [t for t in lines if t[0:2] == last_two]
if not candidates:
break

next_triplet = random.choice(candidates)
last_two = next_triplet[1:3]
output_text.append(next_triplet)

我无法自动执行搜索匹配项并将其存储在新列表中的重复过程。

有什么想法吗？

Answer 1

import random
import sys
import os
import json
outlist = []
file_in = sys.argv[1]
file_ot = str(file_in) + ".ot"
with open(file_in, 'r') as file:
data = file.read().replace('n', '').split()
lines = [data[i:i + 3] for i in range(0, len(data), 1)]
print("nΧωρισμένο Κείμενο σε Λίστα Τριπλετών:n", lines)

triplet = random.choice(lines)
last_two = triplet[1:3]
print("nΕπιλεγμένη Τριπλέτα: n", triplet)
print("nΔύο Τελευταίες Λέξεις Αυτής:n", last_two)
outlist.extend(triplet)
proc_list = lines
# first selected, remove from list
proc_list.remove(triplet)
n = 0
while True:
n += 1
print("nΕπανάληψη {0}n".format(n))
if proc_list == 0:
print("a")
break
random.shuffle(proc_list)

candidates = []

for element in proc_list:
if element[:2] == last_two:
candidates.append(element)

print(candidates)
if not candidates:
print("b")
break


if len(outlist) >= 200:   
print("c")
break

triplet = random.choice(candidates)
outlist.append(triplet[-1])
proc_list.remove(triplet)

last_two = triplet[1:3]
print(outlist)
with open(file_ot, 'w') as f:
f.write(json.dumps(outlist, indent=10))
print(" ".join(outlist))

Answer 2

您可以使用递归函数(更改了代码的某些部分，检查注释(：

import random
#adding ["is", "an", "experiment"] to check if it works (no other triplet were present that satisfy the condition)
lines = [['Python', 'is', 'an'], ['interpreted,', 'high-level', 'and'], ['general-purpose', 'programming', 'language.'], ["Python's", 'design', 'philosophy'], ['emphasizes', 'code', 'readability'], ['with', 'its', 'notable'], ['use', 'of', 'significant'], ['whitespace.', 'Its', 'language'], ['constructs', 'and', 'object-oriented'], ['approach', 'aim', 'to'], ['help', 'programmers', 'write'], ['clear,', 'logical', 'code'], ['for', 'small', 'and'], ['large-scale', 'projects.'], ["is", "an", "experiment"]]
first_triplet = ['Python', 'is', 'an'] # random.choice(lines)

def appendNextTriplet(output_text, lines):
if len(output_text) >= 200:
return output_text
candidates = [t for t in lines if t[:2] == output_text[-2:]]
if not candidates:
return output_text
next_triplet = random.choice(candidates)
output_text += next_triplet # changed from append to concatenation, it was not correct
return appendNextTriplet(output_text, lines)
print(appendNextTriplet(first_triplet, lines)) # ['Python', 'is', 'an', 'is', 'an', 'experiment']