打印带有连续字母数字的字符串

Question

I'm trying to print a string which has consecutive amounts of each of the letters, so the string should print "aaabaaccc". Can anyone tell me where i'm going wrong please as i'm only a beginner in python

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

g = ''

for f in h:

    g = g + f

Answer 1

You can use a Python list comprehension to do this which avoids string concatenation.

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

This will print:

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 repeats String n times.

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

You are trying to unpack your tuple (two items) into one, which you can't do.

Try this instead:

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

Answer 5

Python let's you "multiply" strings by numbers:

>>> 'a' * 5
'aaaaa'

>>> 'ab' * 3
'ababab'

So you can loop over the list, "multiply" the strings by the numbers, and build the string that way.

>>> 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])