Question

在字符方式交换两个字符串时出错...

代码：

#Exchange names using 2-D array of characters

names=["akshay","parag","raman","srinivas","gopal","rajesh"]

      print "Original:%s %s\n" % (names[2],names[3])

for i in range(0,10,1):
      t=names[2][i]
      names[2][i]=names[3][i]
      names[3][i]=t
      print "New:%s %s\n" % (names[2],names[3])

TypeError: 'str' object does not support item assignment

任何解决方案？

Answer 1

names是一个字符串列表。因此，当您执行names[2][someIndex]时，您正在访问该字符串的单个字符。所以当你names[2][i] = names[3][i]时，你试图交换两个字符。因为Python字符串是不可变的，所以这不起作用。

除此之外，您还尝试访问不存在的该字符串的索引。例如，names[2]为raman，因此长度为5个字符。由于您将i循环到9，因此您最终会尝试访问'raman'[5]，'raman'[6]，...，'raman'[9]这些都会引发IndexError。< / p>

如果要在列表中交换两个字符串，只需将字符串作为一个整体进行交换：

>>> names = ["akshay","parag","raman","srinivas","gopal","rajesh"]
>>> t = names[2]
>>> names[2] = names[3]
>>> names[3] = t

或者简而言之，没有临时变量：

>>> names[2], names[3] = names[3], names[2]

Answer 2

字符串在Python中是不可变的，您可以在这里使用bytearray：

names = map(bytearray, names)
print "Original:%s %s\n" % (names[2],names[3])
for i in range(10):
    try:
        #That's how we swap items in Python.
        names[2][i],  names[3][i] = names[3][i], names[2][i]
    except IndexError:
        #'raman' will raise error for index 5, so we need to catch that here
        #Do someting here for unequal strings or break
        break

print "New:%s %s\n" % (names[2],names[3])

您无法更新字符串，您可以使用字符串方法或切片等创建新字符串。

<强>更新

names = map(bytearray, names)
print "Original:%s %s\n" % (names[2],names[3])
s = sorted((names[2], names[3]), key=len)
for i in range(len(s[1])):
    try:
        # Here in case of unequal length string, s[0] is the shorter string
        # and s[1] is the longer one.
        s[0][i], s[1][i] = s[1][i], s[0][i]
    except IndexError:
        for _ in range(i, len(s[1])):
            #remove the items from the longer string and append them to the shorter one.
            s[0].append(s[1].pop(i))            

print "New:%s %s\n" % (names[2],names[3])

另请注意，bytearray只有在使用ASCII字符时才可以使用，不要将其与非ASCII数据一起使用。

Answer 3

这会正确地将names[2]和names[3]换成一行：

names[2], names[3] = names[3], names[2] # they are swapped - a,b = b,a

Answer 4

如果你的每个字符串至少有10个字符，你可以完全没有循环：

names[2], names[3] = names[3][:10] + names[2][10:], names[2][:10] + names[3][10:]

这可以同时完成两项任务。在此之后，names[2]将包含names[3]的前10个字母和位置10的names[2]字母。同样适用于names[3]。

交换两个字符串时出错

4 个答案: