回答編集履歴
3
余ったものも平均を求める処理を追加
answer
CHANGED
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@@ -21,4 +21,11 @@
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21
21
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```Python
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22
22
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def n_mean(ls, n):
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23
23
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return [sum(ls[i:i+n])/n for i in range(0,len(ls),n)]
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24
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+
```
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25
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n 個に満たない余ったものを余った個数で平均して
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26
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n_mean([1,2,3,4,5,6,7,8,9,10], 3) を [2.0, 5.0, 8.0, 10.0] としたい場合は
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27
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```Python
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28
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def n_mean(ls, n):
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29
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ln = len(ls)
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return [sum(ls[i:i+n])/(n if i+n <= ln else ln-i) for i in range(0, ln, n)]
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24
31
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```
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2
コードを戻して処理の差分を追記
answer
CHANGED
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@@ -1,8 +1,7 @@
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1
1
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素直に実装すると
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2
2
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```Python
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3
3
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def n_mean(ls, n):
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4
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-
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4
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return [sum(ls[i-n+1:i+1])/n for i in range(n-1,len(ls),n)]
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5
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return [sum(ls[i:i+n])/n for i in range(0,len(ls),n)] # ppaul さんのご指摘を反映
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6
5
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7
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n_mean([1,2,3,4,5,6,7,8,9], 3) # [2.0, 5.0, 8.0]
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8
7
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@@ -10,4 +9,16 @@
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10
9
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amp_list = n_mean(amp_list, 3)
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11
10
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plt.scatter(time_list,amp_list)
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12
11
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plt.show()
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12
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```
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13
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n 個の平均を取る仕様が曖昧なのでどちらがいいかわかりませんが n 個ずつにした余りを無視して
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14
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n_mean([1,2,3,4,5,6,7,8,9,10], 3) を [2.0, 5.0, 8.0] としたい場合は
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15
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```Python
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16
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def n_mean(ls, n):
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17
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return [sum(ls[i-n+1:i+1])/n for i in range(n-1,len(ls),n)]
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18
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```
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19
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n 個に満たないものも n で割った平均値にして
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20
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n_mean([1,2,3,4,5,6,7,8,9,10], 3) を [2.0, 5.0, 8.0, 3.333...] としたい場合は
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21
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```Python
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22
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def n_mean(ls, n):
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23
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return [sum(ls[i:i+n])/n for i in range(0,len(ls),n)]
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13
24
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```
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1
ppaul さんのご指摘を反映
answer
CHANGED
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@@ -1,7 +1,8 @@
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1
1
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素直に実装すると
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2
2
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```Python
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3
3
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def n_mean(ls, n):
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4
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-
return [sum(ls[i-n+1:i+1])/n for i in range(n-1,len(ls),n)]
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4
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#return [sum(ls[i-n+1:i+1])/n for i in range(n-1,len(ls),n)]
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5
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return [sum(ls[i:i+n])/n for i in range(0,len(ls),n)] # ppaul さんのご指摘を反映
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5
6
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6
7
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n_mean([1,2,3,4,5,6,7,8,9], 3) # [2.0, 5.0, 8.0]
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7
8
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