如何将字符串与字符串列表中的每个元素连接起来?

时间:2017-09-08 20:06:35

标签: python string list

我想将字符串与给定字符串列表中的每个元素连接起来:

my_variable = 'Blue'
my_list = ['House', 'Paint', 'Cloth']

如何将my_variable连接到my_list中的每个元素,以便我有一个新列表

my_new_list = ['Blue_House', 'Blue_Paint', 'Blue_Cloth']

4 个答案:

答案 0 :(得分:2)

或使用列表理解:

print([my_variable + '_' + e for e in my_list])
['Blue_House', 'Blue_Paint', 'Blue_Cloth']

答案 1 :(得分:1)

["_".join([my_variable, i]) for i in my_list]
# ['Blue_House', 'Blue_Paint', 'Blue_Cloth']

答案 2 :(得分:0)

创建一个新列表并遍历原始列表,将原始列表的条目附加到新列表中,并将“Blue_”添加到前面,带有一个简单的+

my_variable = 'Blue'
my_list = ['House', 'Paint', 'Cloth']

new_list = []
for entry in my_list:
    new_list.append(my_variable + "_" + str(entry))

print new_list
>>>>['Blue_House', 'Blue_Paint', 'Blue_Cloth']

答案 3 :(得分:-1)

my_variable1 = 'Blue'
my_variable2 = 'Red'
my_variable3 = 'Black'
my_list = ['House', 'Paint', 'Cloth']

def mergetwo(a, b, s):
    return s.join((a,b))

def prefix(item, pre):
    return lambda sep: mergetwo(pre,item,sep)

def suffix(item, suf):
    return prefix(suf, item)

def mergewith(alist):
    return lambda position: lambda sep: lambda var: list(map(lambda item: position(item, var)(sep), alist))

def addprefix_underscore_mylist(var):
    return mergewith(my_list)(prefix)('_')(var)

def addsuffix_underscore_mylist(var):
    return mergewith(my_list)(suffix)('_')(var)

print(addprefix_underscore_mylist(my_variable1))
print(addprefix_underscore_mylist(my_variable2))
print(addprefix_underscore_mylist(my_variable3))

print(addsuffix_underscore_mylist(my_variable1))
print(addsuffix_underscore_mylist(my_variable2))
print(addsuffix_underscore_mylist(my_variable3))
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