在列表中拆分单词并迭代字典

Question

我有一些代码，我在一个文本中收到一串语言。

我的目标是将此输入转换为列表，并在字典中遍历该列表，以用作值输出的键。我将这个输出发送到一个列表以供稍后使用。

我期望的输出是[57,20,22,52,60…]等]但目前，我收到

[57,没有,没有,没有,没有,没有,没有…等)

我的第一个输出是正确的，但之后，它似乎没有找到字典中正确的值。

下面的代码。

l_languages = []
language_dict = { 'Afrikaans' : 57, 'Arabic' : 20, 'Assistive communication' : 21, 'AUSLAN' : 22, 'Bosnian' : 52,'Burmese' : 60, 'Cantonese' : 23, 'Croation' : 54, 'Dutch' : 50,'French' : 24, 'German' : 25, 'Greek' : 26,'Hindi' : 27, 'Indigenous Australian' : 310, 'Indonesian' : 56, 'Italian' : 28, 'Japanese' : 62, 'Korean' : 48, 'Mandarin' : 29, 'Nepali' : 55, 'Polish' : 30}
data = "Afrikaans, Arabic, Assistive communication, AUSLAN, Bosnian, Burmese, Cantonese, Croation, Dutch"

language_list = data.split(',')
for language in language_list:
id = language_dict.get(language)
l_languages.append(id)

print(l_languages)
current output = [57, None, None, None, None, None, None....etc]

Answer 1

您忽略了语言列表中的空白。您应该删除前后的空白，并访问您的字典。

如果你只是在'，'处分割列表，那么接下来的每个语言前面都有一个前导空白。只是不是第一个，它解释了您当前的输出

Answer 2

看看你的language_list。它有前导空格。您需要在每个元素上调用strip()，然后得到预期的结果

l_languages = []
language_dict = { 'Afrikaans' : 57, 'Arabic' : 20, 'Assistive communication' : 21, 'AUSLAN' : 22, 'Bosnian' : 52,'Burmese' : 60, 'Cantonese' : 23, 'Croation' : 54, 'Dutch' : 50,'French' : 24, 'German' : 25, 'Greek' : 26,'Hindi' : 27, 'Indigenous Australian' : 310, 'Indonesian' : 56, 'Italian' : 28, 'Japanese' : 62, 'Korean' : 48, 'Mandarin' : 29, 'Nepali' : 55, 'Polish' : 30}
data = "Afrikaans, Arabic, Assistive communication, AUSLAN, Bosnian, Burmese, Cantonese, Croation, Dutch"
language_list = data.split(',')
print(language_list)
for language in language_list:
val = language_dict.get(language.strip())
l_languages.append(val)

print(l_languages)
['Afrikaans', ' Arabic', ' Assistive communication', ' AUSLAN', ' Bosnian', ' Burmese', ' Cantonese', ' Croation', ' Dutch'] # list with leading spaces
[57, 20, 21, 22, 52, 60, 23, 54, 50] # right result

Answer 3

l_languages = []
language_dict = { 'Afrikaans' : 57, 'Arabic' : 20, 'Assistive communication' : 21, 'AUSLAN' : 22, 'Bosnian' : 52,'Burmese' : 60, 'Cantonese' : 23, 'Croation' : 54, 'Dutch' : 50,'French' : 24, 'German' : 25, 'Greek' : 26,'Hindi' : 27, 'Indigenous Australian' : 310, 'Indonesian' : 56, 'Italian' : 28, 'Japanese' : 62, 'Korean' : 48, 'Mandarin' : 29, 'Nepali' : 55, 'Polish' : 30}
data = "Afrikaans, Arabic, Assistive communication, AUSLAN, Bosnian, Burmese, Cantonese, Croation, Dutch"
language_list=[x.strip() for x in data.split(',')]
for language in language_list:
id = language_dict.get(language)
l_languages.append(id)
#output
[57, 20, 21, 22, 52, 60, 23, 54, 50]

Answer 4

最简单的方法

#Devil
language_dict = { 'Afrikaans' : 57, 'Arabic' : 20, 'Assistive communication' : 21,
'AUSLAN' : 22, 'Bosnian' : 52,'Burmese' : 60, 'Cantonese' : 23,
'Croation' : 54, 'Dutch' : 50,'French' : 24, 'German' : 25, 'Greek' : 26,
'Hindi' : 27, 'Indigenous Australian' : 310, 'Indonesian' : 56, 'Italian' : 28,
'Japanese' : 62, 'Korean' : 48, 'Mandarin' : 29, 'Nepali' : 55, 'Polish' : 30}
data = "Afrikaans, Arabic, Assistive communication, AUSLAN, Bosnian, Burmese, Cantonese, Croation, Dutch"
data_list = data.split(",") #split the data
data_list = [d.strip() for d in data_list] #remove white space
l_languages = [language_dict[z] for z in data_list] #find the value using key
print(data_list)
print(l_languages)