回答編集履歴
5
追記
test
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@@ -79,3 +79,7 @@
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[ 0., 2., 1000.]])
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```
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最後の方法以外はin-placeで処理するので、もとのaを残して別途bを作りたければ配列のコピー操作をする必要があります。
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4
追記
test
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@@ -15,6 +15,16 @@
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15
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```python
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a[np.eye(3, dtype=bool)] = 1000
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```
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あるいはこんな。
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```python
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a + 1000*np.eye(3)
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```
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@@ -58,4 +68,14 @@
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[ 0, 2, 1000]])
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>>> a = np.array([[0,1,0],[1,0,2],[0,2,0]])
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>>> a + 1000*np.eye(3)
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array([[1000., 1., 0.],
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[ 1., 1000., 2.],
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[ 0., 2., 1000.]])
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61
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```
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3
追記
test
CHANGED
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@@ -20,7 +20,11 @@
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+
実行例
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```python
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>>> import numpy as np
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>>> a = np.array([[0,1,0],[1,0,2],[0,2,0]])
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2
追記
test
CHANGED
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@@ -17,3 +17,41 @@
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17
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a[np.eye(3, dtype=bool)] = 1000
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```
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```python
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>>> a = np.array([[0,1,0],[1,0,2],[0,2,0]])
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>>> a
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array([[0, 1, 0],
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[1, 0, 2],
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[0, 2, 0]])
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>>> a[0,0] = a[1,1] = a[2,2] = 1000
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>>> a
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array([[1000, 1, 0],
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[ 1, 1000, 2],
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[ 0, 2, 1000]])
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>>> a = np.array([[0,1,0],[1,0,2],[0,2,0]])
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>>> a[np.eye(3, dtype=bool)] = 1000
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>>> a
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array([[1000, 1, 0],
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[ 1, 1000, 2],
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[ 0, 2, 1000]])
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```
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1
追記
test
CHANGED
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@@ -5,3 +5,15 @@
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5
5
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a[0,0] = a[1,1] = a[2,2] = 1000
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6
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```
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8
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11
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ちょっと凝ってみると、こうした方がスマートかもしれません。
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```python
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16
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a[np.eye(3, dtype=bool)] = 1000
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```
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