我有一个字符串（长度未定），我想复制很多次，从字符数组（长度未定）中一次替换一个字符。
假设我有这个字符串：'aa'
而这个数组：['a', 'b', 'c', 'd']
在一些神奇的for循环之后会有一个数组，如：['aa', 'ab', 'ac', 'ad', 'ba', 'bb' ... 'dc', 'dd ']
你会怎么做？我尝试了使用三个 for 循环的东西，但我似乎无法得到它。
编辑
对字符串的依赖如下：
假设字符串是：'ba'
那么输出应该是：['ba', 'bb', 'bc', 'bd', 'ca' ... 'dd']
【问题讨论】：
如果结果数组中的字符串顺序无关紧要，并且初始字符串中的所有字符都在替换数组中，则：
#!/usr/bin/env python
from itertools import product
def allreplacements(seed, replacement_chars):
assert all(c in replacement_chars for c in seed)
for aset in product(replacement_chars, repeat=len(seed)):
yield ''.join(aset)
print(list(allreplacements('ba', 'a b c d'.split())))
# ['aa', 'ab', 'ac', 'ad', 'ba', 'bb', 'bc', 'bd', 'ca', 'cb', 'cc',
# 'cd', 'da', 'db', 'dc', 'dd']
这是针对一般情况的解决方案。替换按字典顺序执行：
#!/usr/bin/env python
from itertools import product
def allreplacements(seed, replacement_chars):
"""Generate all possible replacements (with duplicates)."""
masks = list(product(range(2), repeat=len(seed))) # e.g., 00 01 10 11
for subs in product(replacement_chars, repeat=len(seed)):
for mask in masks:
# if mask[i] == 1 then replace seed[i] by subs[i]
yield ''.join(s if m else c for s, m, c in zip(subs, mask, seed))
def del_dups(iterable):
"""Remove duplicates while preserving order.
http://stackoverflow.com/questions/89178/in-python-what-is-the-fastest-algorithm-for-removing-duplicates-from-a-list-so#282589
"""
seen = {}
for item in iterable:
if item not in seen:
seen[item] = True
yield item
print(list(del_dups(allreplacements('ba', 'abcd'))))
print(list(del_dups(allreplacements('ef', 'abcd'))))
# ['ba', 'aa', 'bb', 'ab', 'bc', 'ac', 'bd', 'ad', 'ca', 'cb', 'cc',
# 'cd', 'da', 'db', 'dc', 'dd']
# ['ef', 'ea', 'af', 'aa', 'eb', 'ab', 'ec', 'ac', 'ed', 'ad', 'bf',
# 'ba', 'bb', 'bc', 'bd', 'cf', 'ca', 'cb', 'cc', 'cd', 'df', 'da',
# 'db', 'dc', 'dd']
【讨论】：
print map(''.join,product('abcd',repeat=len('ba')))
您可以通过两种方式使用以下代码：
对于用法 (1)，调用 getStrings() 方法（根据需要多次）。
对于用法 (2)，只要 hasNext() 返回 true，就仅调用 next() 方法。 （实现reset() 方法留给读者作为练习！;-)
package com.so.demos;
import java.util.ArrayList;
import java.util.List;
public class StringsMaker {
private String seed; // string for first value
private char[] options; // allowable characters
private final int LAST_OPTION; // max options index
private int[] indices; // positions of seed chars in options
private int[] work; // positions of next string's chars
private boolean more; // at least one string left
public StringsMaker(String seed, char[] options) {
this.seed = seed;
this.options = options;
LAST_OPTION = options.length - 1;
indices = new int[seed.length()];
for (int i = 0; i < indices.length; ++i) {
char c = seed.charAt(i);
for (int j = 0; j <= LAST_OPTION; ++j) {
if (options[j] == c) {
indices[i] = j;
break;
}
}
}
work = indices.clone();
more = true;
}
// is another string available?
public boolean hasNext() {
return more;
}
// return current string, adjust for next
public String next() {
if (!more) {
throw new IllegalStateException();
}
StringBuffer result = new StringBuffer();
for (int i = 0; i < work.length; ++i) {
result.append(options[work[i]]);
}
int pos = work.length - 1;
while (0 <= pos && work[pos] == LAST_OPTION) {
work[pos] = indices[pos];
--pos;
}
if (0 <= pos) {
++work[pos];
} else {
more = false;
}
return result.toString();
}
// recursively add individual strings to result
private void getString(List<String> result, int position, String prefix) {
if (position == seed.length()) {
result.add(prefix);
} else {
for (int i = indices[position]; i < options.length; ++i) {
getString(result, position + 1, prefix + options[i]);
}
}
}
// get all strings as array
public String[] getStrings() {
List<String> result = new ArrayList<String>();
getString(result, 0, "");
return result.toArray(new String[result.size()]);
}
}
【讨论】：
嗯，两个 for 循环应该可以做到：Python 伪代码--
a = "abcd"
b = "ba"
res = []
for i in a: # i is "a", "b", ...
for j in b: # j is "b", "a"
res.append(i+j) # [ "ab", "bb",...]
return res
[更新：纠正了愚蠢的错字。]
【讨论】：
res.append(i + j)？这是错误的。结果字符串始终以 b 中的字符结尾。
如果字符串和数组不都包含“a”，问题会更清楚。所需的输出不显示对输入字符串的任何依赖。
【讨论】：