如何在python

时间:2017-11-18 11:43:23

标签: python string split

我有一个像'0111111011111100'这样的字符串,我希望将每四个字符分开,这样就可以了:

0111,1110,1100

然后我想用其他值替换那些。

到目前为止,这是我的代码,但它无法正常运行:

4 个答案:

答案 0 :(得分:4)

您可以list-comprehension使用string indexing

s = "0111111011111100"
[s[i:i+4] for i in range(0,len(s),4)]

给出:

['0111', '1110', '1111', '1100']

然后为每个dictionary要翻译的内容定义nibble

d = {'0111': 'a', '1110': 'b', '1111': 'c', '1100': 'd'}

然后你可以将翻译推到list-comp

[d[s[i:i+4]] for i in range(0,len(s),4)]

会给出:

['a', 'b', 'c', 'd']

并最终使用str.join将其重新放回string,使整个转换成为一行:

''.join(d[s[i:i+4]] for i in range(0,len(s),4))

给出:

'abcd'

事实上,这里使用generator表达式,因为它们比list-comprehensions

更有效

答案 1 :(得分:0)

如果您想用字符串替换字符串中的每个四位数组,可以使用dict

line='0111111011111100'
lookup = {'0111': 'a', '1110': 'b', '1100': 'c'} # add all combination here
"".join(lookup.get(line[i:i+4], 'D') for i in range(0, len(line), 4)) # 'D' is default value
Out[18]: 'abDc'

答案 2 :(得分:0)

line='0111111011111100'
# define a dictionary with nibbles as values
dict = { '0111': 'a', '1110': 'b', '1111': 'c', '1100': 'd'}

# define chunk size
n = 4

# use a list comprehension to split the original string into chunks
key_list = [line[i:i+n] for i in range(0, len(line), n)]

# use a generator expression to replace keys in the list with
# their dictionary values and join together
key_list = ' '.join(str(dict.get(value, value)) for value in key_list)
print(key_list)

答案 3 :(得分:0)

这可能对您有所帮助:

str = "0111111011111100"

n = 4

# Create a list from you string with 4 characters in one element of list.
tmp_list = [str[i:i + n] for i in range(0, len(str), n)]
# tmp_list : ['0111', '1110', '1111', '1100']

for n, i in enumerate(tmp_list):
    if tmp_list[n] == "0111":
        tmp_list[n] = "A"
    # elif ....
    # Todo:
    # Populate your if-elif requirements here.   

result_str = ''.join(tmp_list)