Question

从文本文件行读取后，我有一个这样的字符串：

poLCA::poLCA()

我想将其转换为元组。我在做的是：

#Latent class models with one (loglinear independence) to three classes
f=cbind("I_like_to_drink","I_drink_hard_liquor","I_have_drank_in_the_morning","I_have_drank_in_the_work","I_drink_to_get_drunk","I_like_the_taste_of_the_alcohol","I_drink_help_me_sleep","Drinking_interferes_with_my_relationships","I_frequently_visit_bars")~1
    nes1 = poLCA(f,lca1,nclass=1) # log-likelihood: 
    nes2 = poLCA(f,lca1,nclass=2) # log-likelihood: 
    nes3 = poLCA(f,lca1,nclass=3) # log-likelihood: 
    nes4 = poLCA(f,lca1,nclass=4) # log-likelihood:

预计我需要将每个值除以To = "'abc@hcl.com','xyz@hcl.com','accc@infy.com','satya@lab.com','ach@lab.com'"并将其存储到字符串或元组转换后的元组中：

To = (To,)
type(To)  #o/p tuple
len(To)   #1
To
("'abc@hcl.com','xyz@hcl.com','accc@infy.com', 'sat@lab.com','ach@lab.com'",)   # output tuple

Answer 1

使用Abstract Syntax Trees，如下所示 -

>>>import ast
>>>To=ast.literal_eval(To)
>>>print To
>>>('abc@hcl.com', 'xyz@hcl.com', 'accc@infy.com', 'satya@lab.com', 'ach@lab.com')
>>>len(To) 
>>>5

Answer 2

或者如果你不想使用ast，你可以简单地运行：

tuple(word[1:-1] for word in To.split(","))

Answer 3

备注：

您可以通过String
拆分将List转换为,
然后，您可以将List转换为Tuple

<强>代码：

To = "'abc@hcl.com','xyz@hcl.com','accc@infy.com','satya@lab.com','ach@lab.com'" print tuple(To.split(","))

<强>输出：

("'abc@hcl.com'", "'xyz@hcl.com'", "'accc@infy.com'", "'satya@lab.com'", "'ach@lab.com'")

如果您要删除'引号

<强>代码1：

print tuple(elem.strip("'") for elem in To.split(","))

<强>输出1：

('abc@hcl.com', 'xyz@hcl.com', 'accc@infy.com', 'satya@lab.com', 'ach@lab.com')

字符串转换为python中的元组转换

3 个答案: