打印带有连续字母数字的字符串

Question

我正在尝试打印包含每个字母连续数量的字符串，因此该字符串应打印“ aaabaaccc”。 谁能告诉我我要去哪里错了，因为我只是python的初学者

h = [("a", 3), ("b", 1), ("a", 2), ("c", 3)]

g = ''

for f in h:

    g = g + f

Answer 1

您可以使用Python列表推导来避免字符串连接。

print ''.join(letter * count for letter, count in [("a", 3), ("b", 1), ("a", 2), ("c", 3)])

这将打印：

aaabaaccc

Answer 2

h = [("a", 3), ("b", 1), ("a", 2), ("c", 3)]
g = ''
for char, count in h:
    #g = g + f  #cant convert tuple to string implicitly
    g=g+char*count
print(g)

String*n重复String n次。

Answer 3

h = [("a", 3), ("b", 1), ("a", 2), ("c", 3)]

g = ''

for i, j in h:     # For taking the tuples into consideration


    g += i * j

print(g)  # printing outside for loop so you get the final answer (aaabccc)

Answer 4

您正在尝试将元组（两个项目）拆成一个包装，而这是做不到的。

尝试以下方法：

for letter, multiplier in h:
    g += letter * multiplier

Answer 5

Python让您将字符串乘以数字：

>>> 'a' * 5
'aaaaa'

>>> 'ab' * 3
'ababab'

因此，您可以遍历列表，将字符串乘以数字，然后以这种方式构建字符串。

>>> h = [("a", 3), ("b", 1), ("a", 2), ("c", 3)]
>>> g = ''
>>> for letter, number in h:
...     g += letter * number
... 
>>> print(g)
aaabaaccc

Answer 6

如果“ h”总是要这样格式化，那么一个单行代码是：

g = "".join([i[0]*i[1] for i in h])