如何在python中合并两个字符串？

我需要合并字符串以创建一个字符串。例如，需要组合的“hello”字符串包括： [H---]、[-E---]、[-LL-]和[----O]

这是我必须首先让字母出现的当前代码

display_string = "[" + "".join([x if x == letter else "-" for x in word]) + "]"

我该如何制作字符串，例如

---

[H---]、[HE---]、[HELL-]，最后是[HELLO]？

我不知道这是否有助于实现您的愿望，但这是有效的。问题是我们没有真正的字符串替换。在最终组装之前，一个字符列表可能对您更有用

string_list = [
    'H----',
    '-E---',
    '--LL-',
    '----O'
]

word_len = len(string_list[0])
display_string = word_len*'-'
for word in string_list:
    for i in range(word_len):
        if word[i].isalpha():
            display_string = display_string[:i] + word[i] + display_string[i+1:]
    print '[' + display_string + ']'

---

好吧，我想我明白你的意思了

我们定义一个像“hello”这样的词，然后我们迭代“hello”中所有唯一的字母，得到您的

display\u字符串。那么你的问题是如何将它们合并在一起

word = "HELLO"
all_display_strings = ["[" + "".join([x if x == letter else "-" for x in word]) + "]" for letter in word]
# now merge the list of lists
length = len(all_display_strings[0])-2 # -2 for the opening and closing bracket
merged_display_strings = ['-']*length

for i in range(0, length):
        for j in range(0, len(all_display_strings)):
            if (all_display_strings[j][i+1] != '-'):
                merged_display_strings[i] = all_display_strings[j][i+1]


print(''.join(merged_display_strings))

假设您的字符串位于列表
所有\u字符串中

all_strings = ["[H----]", "[-E---]", "[--LL-]", "[----O]"]
# clean them
all_strings = map(lambda x:x.strip()[1:-1], all_strings)

# Let's create a list for the characters in the final string(for easy mutating)
final_string = ['' for _ in xrange(max(map(len, all_strings)))]

for single in all_strings:
    for i, char in enumerate(single):
        if char != '-': final_string[i]=char

print "[" + ''.join(final_string) + "]"

你能解释一下字母是什么吗？
strings = 'h----', '-e---', '--ll-', '----o'
letterSoups = [ [ (p, c) for (p, c) in enumerate(s) if c != '-' ]
    for s in strings ]
result = list('-----')
for soup in letterSoups:
  for p, c in soup:
    result[p] = c
print ''.join(result)

s = "[H----], [-E---], [--LL-],  [----O]"

print(''.join(re.findall(r'\w', s)))

HELLO