回答編集履歴
4
追記
answer
CHANGED
@@ -57,5 +57,25 @@
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57
57
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break
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58
58
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59
59
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dst[-1].append(i)
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60
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-
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60
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+
```
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61
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+
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62
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+
さらに追記
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63
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+
---
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64
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+
普通に読み辛いですし効率面も特段良くありませんが、個人的にはこういうのが好きです。
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65
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+
おまけです。
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66
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+
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67
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+
```Python
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68
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+
indice_1 = (
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69
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+
i for i, e in enumerate(src)
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70
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+
if e == 1
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71
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+
)
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72
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+
dst = [
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73
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+
[idx, *map(operator.itemgetter(0), zip(
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74
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+
itertools.count(idx+1),
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75
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+
itertools.takewhile(
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76
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+
lambda e: e == 2, src[idx+1:]
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77
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)
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78
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+
))]
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79
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+
for idx in indice_1
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80
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+
]
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61
81
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```
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3
追記
answer
CHANGED
@@ -38,4 +38,24 @@
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38
38
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while i < len(src) and src[i] == 2:
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39
39
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dst[-1].append(i)
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40
40
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i += 1
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41
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+
```
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42
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+
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43
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+
for文を使うなら二回に分けた方が無駄が少ないように思います。
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44
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+
```Python
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45
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+
dst = []
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46
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+
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47
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+
it = (
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48
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+
i for i, e in enumerate(src)
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49
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+
if e == 1
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50
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+
)
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51
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+
for index_1 in it:
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52
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+
dst.append([index_1])
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53
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+
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54
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+
idx = index_1 + 1
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55
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+
for i, e in enumerate(src[idx:], start=idx):
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56
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if e != 2:
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57
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+
break
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58
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+
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59
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+
dst[-1].append(i)
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60
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+
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41
61
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```
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2
追記
answer
CHANGED
@@ -18,4 +18,24 @@
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18
18
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**実行結果** [Wandbox](https://wandbox.org/permlink/SVnsvzydmFuJXlyh)
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19
19
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```
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20
20
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[[6], [9, 10, 11]]
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21
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+
```
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22
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+
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23
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+
コメントを受けて
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24
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+
---
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25
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+
例えば、こんなふうに。
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26
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+
```Python
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27
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+
dst = []
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28
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+
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29
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i = 0
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30
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+
while i < len(src):
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31
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if src[i] != 1:
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32
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i += 1
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33
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continue
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34
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+
|
35
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+
dst.append([i])
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36
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+
i += 1
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37
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+
|
38
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+
while i < len(src) and src[i] == 2:
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39
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+
dst[-1].append(i)
|
40
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+
i += 1
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21
41
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```
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1
追記
answer
CHANGED
@@ -3,7 +3,9 @@
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3
3
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import re
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4
4
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5
5
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6
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-
src = [4,
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6
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+
src = [4,4,2,4,5,3,1,3,4,1,2,2,3]
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7
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+
assert all(len(str(e)) == 1 for e in src) # srcの全要素は1桁
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8
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+
|
7
9
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dst = [
|
8
10
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[*range(m.start(), m.end())]
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9
11
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for m in re.finditer(r'12*', ''.join(map(str, src)))
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@@ -13,9 +15,7 @@
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13
15
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print(dst)
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14
16
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```
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15
17
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16
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-
**実行結果** [Wandbox](https://wandbox.org/permlink/
|
18
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+
**実行結果** [Wandbox](https://wandbox.org/permlink/SVnsvzydmFuJXlyh)
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17
19
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```
|
18
20
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[[6], [9, 10, 11]]
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19
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-
```
|
21
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+
```
|
20
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-
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21
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-
srcの全ての要素が1桁の整数との前提です。
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