借助字典更正字符串中单词的拼写？答案

【问题标题】：Correct the spellings of words in string with the help of a dictionary?借助字典更正字符串中单词的拼写？
【发布时间】：2019-01-08 12:01:00
【问题描述】：

我有一个输入列表，其中 REJECTED 的拼写写为 RE1ECTED，而 DATE 的拼写写为 OATE。我需要更正拼写并得到一个输出列表：

output_list = ["REJECTED 00000 00000","DATE TIME =06/27/18 21 37","NACH NE NO XY"]

我能够单独获得更正的列表，但不像上面显示的串联。我可以期待一些建议吗？

input_list = ["=RE1ECTED 00000 00000",'OATE TIME =06/27/18 21 37','NACH NE NO xy']
Reg_dictionary = {"REJECTED" : ["=RE"],"DATE" : ["OA","DA"] }
for key,value in Reg_dictionary.items():
    temp = [key, value]
    for each_value in value:
        for string_list in input_list:
            count  = -1
            # print (count_number_of_string)
            each_in_string  = string_list.split(" ")
            for each_word in each_in_string:
                count = count + 1
                if each_value in each_word:
                    # print (count)
                    # print (string_list)
                    # print(each_word)
                    # print (key)
                    # print(each_in_string)
                    (each_in_string[count]) =key
                    print(each_in_string)

print(each_in_string) 的输出是这样的：

['REJECTED', '00000', '00000']

['DATE', 'TIME', '=06/27/18', '21', '37']

预期输出是：

output_list = ["REJECTED 00000 00000","DATE TIME =06/27/18 21 37","NACH NE NO XY"]

【问题讨论】：

标签： python regex string

【解决方案1】：

您必须将列表中的元素连接在一起，然后将列表连接在一起。您可以使用空格作为元素之间的分隔符连接列表中的元素，如下所示：' '.join(your_list)，这将为您提供字符串。然后只需将 2 个字符串与 [string_1 + ' ' + string_2] 连接在一起。应该这样做:)

【讨论】：

【解决方案2】：

你可以这样做：

from functools import reduce 

input_list = ["=RE1ECTED 00000 00000",'OATE TIME =06/27/18 21 37','NACH NE NO xy']
Reg_dictionary = {"REJECTED" : ["=RE"],"DATE" : ["OA","DA"] }
l = []
for key,value in Reg_dictionary.items():
    temp = [key, value]
    for each_value in value:
        for string_list in input_list:
            count  = -1
            # print (count_number_of_string)
            each_in_string  = string_list.split(" ")
            for each_word in each_in_string:
                count = count + 1
                if each_value in each_word:
                    # print (count)
                    # print (string_list)
                    # print(each_word)
                    # print (key)
                    # print(each_in_string)
                    (each_in_string[count]) =key
                    #print(each_in_string)
                    l.append(reduce(lambda x,y: x+' '+y, each_in_string)) # added this line
print(l)

【讨论】：

嗨@Rodwan Bakkar 感谢您的帮助，我需要一个包含所有三个字符串的输出。 output_list = ["拒绝 00000 00000","日期时间 =06/27/18 21 37","NACH NE NO XY"]
要做到这一点，根据您的方法，您必须添加 Reg_dictionary = {"REJECTED" : ["=RE"],"DATE" : ["OA","DA"], "NACH" : ["NA"] } 之类的内容，但这不是正确的方法，您的代码有点混乱。