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print_find.py
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print_find.py
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import re
mode = input("请输入模式，输入0为输入1个字母查找音标，输入1为输入1个音标查找字母\n"
"输入2为输入2及以上个字母查找音标，输入3为输入2及以上个音标查找字母：")
if mode == '0':
g_p = input("输入所需查找的1个字母：")
elif mode == '1':
g_p = input("输入所需查找的1个音标：")
elif mode == '2':
g_p = input("输入所需查找的n个字母：")
else:
g_p = input("输入所需查找的n个音标：")
with open('g2p_coca_ipa_withoutstress.txt', 'r', encoding='utf-8') as f: # gbk编码会读成别的字符
lines = f.read().split('\n')
#
# 重定向print到文件
import sys
file_path = 'print_find.txt'
sys.stdout = open(file_path, "a",encoding='utf-8')#不能放在最前边，否则input()的输出就没了
#
# European 转小写
#
#
# g_p = "i"
dict_print = {}
for line in lines:
ll = line.split()
#
# for i in ll[1:]:
if mode == '2' or mode == '3':
g_p_len = len(g_p)
for i in range(1, len(ll)):
substr_grapheme = ""
substr_phoneme = ""
substr_total = ""
#
if mode == '2':
if g_p[0] not in ll[i].split("}")[0].replace('|', ''):
continue#第一个必须存在，防止出现lea-please p}p l}l e|a}i这样的情况
for j in range(i, len(ll)):
substr_grapheme += ll[j].split("}")[0].replace('|', '')
substr_phoneme += ll[j].split("}")[1]
substr_total += " " + ll[j]
if g_p in substr_grapheme:
key = substr_phoneme
if key not in dict_print:
dict_print[key] = []
dict_print[key].append(ll[0] + substr_total)
break
# 需要改这句，否则会出现league g|u占了两个长度
len_space = j - i + 1
if len_space == g_p_len:#这个条件必须后判断，有|的存在，len_space的终止长度是不确定的
break
# 更简单的思路或许是将e|a}i 这样的情况分成e}i和a}i e和a都对应一个e|a}i，最后去重即可；这样就可以直接子字符串判断了，不需要考虑多变的长度问题。。。
#
elif mode == '3':
if g_p[0] not in ll[i].split("}")[1].replace('|', ''):
continue # 第一个必须存在，防止出现lea-please p}p l}l e|a}i这样的情况
for j in range(i, len(ll)):
substr_grapheme += ll[j].split("}")[0].replace('|', '')
substr_phoneme += ll[j].split("}")[1]
substr_total += " " + ll[j]
if g_p in substr_phoneme:
key = substr_grapheme
if key not in dict_print:
dict_print[key] = []
dict_print[key].append(ll[0] + substr_total)
break
# 需要改这句，否则会出现league g|u占了两个长度
len_space = j - i + 1
if len_space == g_p_len: # 这个条件必须后判断，有|的存在，len_space的终止长度是不确定的
break
# offer o}ɔ f|f}f e|r}ɚ
else:
for i in range(1, len(ll)):# 这里同一个单词内的多个g_p分成了多行append进
#
#
if mode == '0' or mode == '1':
grapheme_phoneme = ll[i].split("}")
# 0 1统一
if g_p in grapheme_phoneme[int(mode)]:
key = grapheme_phoneme[1 - int(mode)]
if key not in dict_print:
dict_print[key] = []
dict_print[key].append(line + " | " + ll[i])# 同一个单词里有多个g_p的话，就分不清是哪个
# print(ll[0] + " " + i)
# if mode == '2':
# grapheme_phoneme = ll[i].split("}")
# if i != len(ll) - 1:
# grapheme_phoneme_2 = ll[i + 1].split("}")
# else:
# grapheme_phoneme_2 = ['@','@']
# # grapheme_phoneme_2 = ''
# app = ''
# if g_p[0] in grapheme_phoneme[0][-1] and g_p[1] in grapheme_phoneme_2[0][0]:
# key = grapheme_phoneme[1]+grapheme_phoneme_2[1]
# if key not in dict_print:
# dict_print[key] = []
# dict_print[key].append(ll[0] + " " + ll[i] + " " + ll[i + 1])
# # print(ll[0] + " " + ll[i] + " " + ll[i + 1])
# if g_p in grapheme_phoneme[0].replace('|',''):
# key = grapheme_phoneme[1]
# if key not in dict_print:
# dict_print[key] = []
# dict_print[key].append(ll[0] + " " + ll[i])
#
print("\n\n\n@@")
# 求总数量，计算百分比
# 正则过滤
# re_str = "^se"
try:
if re_str:
print(1)
except:
re_str = ""
# re_str = ""
len_total = 0
for k, v_list in dict_print.items():
# print(k,">>",v)
for line in v_list[:]:
if re_str and (re.match(re_str, line.split()[0]) == None):
v_list.remove(line)
len_total += len(v_list)
#
dict_print_list = sorted(dict_print.items(), key=lambda kv: len(kv[1]), reverse=True)
for tuple in dict_print_list:
print("---------------------------")
# print(tuple[0])
print(tuple[0],len(tuple[1]),len(tuple[1])/len_total)
print("+++++++++++++++++++++++++++")
for i in tuple[1]:
print(i)
#r}.{1,3}e}.{1,3}o}.{1,2}