回答編集履歴
1
ソースの追記
answer
CHANGED
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@@ -24,4 +24,48 @@
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24
24
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d
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25
25
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e
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26
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f
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27
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```
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28
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29
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30
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2016/8/28追記：
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31
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0or1のある・ないだけ知れば良いのであればディクショナリ型を使うまでもなさそうですね。
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32
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set型にしてitem名だけ保持させて、呼び出し時に判定します。
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33
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34
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```python
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35
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table = []
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36
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37
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table.append(["a","c","d"])
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38
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table.append(["a","b"])
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39
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table.append(["a","d"])
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40
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41
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print "a" in table[0]
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print "b" in table[0]
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print "c" in table[0]
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print "d" in table[0]
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45
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46
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print "a" in table[1]
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47
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print "b" in table[1]
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48
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print "c" in table[1]
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49
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print "d" in table[1]
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50
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51
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print "a" in table[2]
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print "b" in table[2]
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53
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print "c" in table[2]
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print "d" in table[2]
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55
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```
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56
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57
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実行結果
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58
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```
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59
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True
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60
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False
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61
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True
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62
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True
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63
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True
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64
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True
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65
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False
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66
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False
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True
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68
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False
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69
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False
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70
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True
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27
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```
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