Question

reads = { '1': 'A', '2': 'B', '3': 'C', '4': 'D', '5': 'E', '6': 'F' }`
readOrder = ['1', '2', '3', '4', '5', '6']`

如何在readOrder中迭代字符串：

a = []
for i in readOrder():
    a.append(reads[i],reads[i+1])

''.join(a)

print a

所以我可以得到：

'ABBCCDDEEF'

这种方式有效，但我必须有一个简单的方法来使用for循环：

a = [reads[0] + reads[1], reads[1] + reads[2], 
     reads[2] + reads[3], reads[3] + reads[4]]
print ''.join(a)

Answer 1

a = []
for i,key in enumerate(readOrder):
 a.append(reads[key])
 if i != 0 and i != len(readOrder) - 1:
   a.append(reads[key])
''.join(a)

Answer 2

也许是这样的？（修改为字符串）

#!/usr/local/cpython-2.7/bin/python

reads = { '1': 'A', '2': 'B', '3': 'C', '4': 'D', '5': 'E', '6': 'F' }

readOrder = ['1', '2', '3', '4', '5', '6']

a = []
for indexno, index_str in enumerate(readOrder[:-1]):
    index_str_p1 = readOrder[indexno + 1]
    a.append(reads[index_str])
    a.append(reads[index_str_p1])

print a
print ''.join(a)

在python中迭代列表中的字符串

2 个答案: