我有一个两列numpy数组。我想通过第2列的每一行，并取每组2个数字（9.6-0,19.13-9.6等）之间的差异。如果差异> 15，我想为两列插入一行0。我真的只需要在第一列中结束值（我只需要第二列来确定将0放在哪里），所以如果它更容易将它们拆分就可以了。
这是我的输入数组：
[[0.00 0.00]
[1.85 9.60]
[2.73 19.13]
[0.30 28.70]
[2.64 38.25]
[2.29 47.77]
[2.01 57.28]
[2.61 66.82]
[2.20 76.33]
[2.49 85.85]
[2.55 104.90]
[2.65 114.47]
[1.79 123.98]
[2.86 133.55]]
它应该变成：
[[0.00 0.00]
[1.85 9.60]
[2.73 19.13]
[0.30 28.70]
[2.64 38.25]
[2.29 47.77]
[2.01 57.28]
[2.61 66.82]
[2.20 76.33]
[2.49 85.85]
[0.00 0.00]
[2.55 104.90]
[2.65 114.47]
[1.79 123.98]
[2.86 133.55]]
答案 0 :(得分：7)
您可以使用ediff1d中的argmax，insert和numpy进行一次性投注：
np.insert(arr, np.argmax(np.append(False, np.ediff1d(arr[:,1])>15)), 0, axis=0)
#array([[ 0. , 0. ],
# [ 1.85, 9.6 ],
# [ 2.73, 19.13],
# [ 0.3 , 28.7 ],
# [ 2.64, 38.25],
# [ 2.29, 47.77],
# [ 2.01, 57.28],
# [ 2.61, 66.82],
# [ 2.2 , 76.33],
# [ 2.49, 85.85],
# [ 0. , 0. ],
# [ 2.55, 104.9 ],
# [ 2.65, 114.47],
# [ 1.79, 123.98],
# [ 2.86, 133.55]])
答案 1 :(得分：6)
假设A为输入数组，这里是一个基于零初始化的矢量化方法 -
# Get indices at which such diff>15 occur
cut_idx = np.where(np.diff(A[:,1]) > 15)[0]
# Initiaize output array
out = np.zeros((A.shape[0]+len(cut_idx),2),dtype=A.dtype)
# Get row indices in the output array at which rows from A are to be inserted.
# In other words, avoid rows to be kept as zeros. Finally, insert rows from A.
idx = ~np.in1d(np.arange(out.shape[0]),cut_idx + np.arange(1,len(cut_idx)+1))
out[idx] = A
示例输入，输出 -
In [50]: A # Different from the one posted in question to show variety
Out[50]:
array([[ 0. , 0. ],
[ 1.85, 0.6 ],
[ 2.73, 19.13],
[ 2.2 , 76.33],
[ 2.49, 85.85],
[ 2.55, 104.9 ],
[ 2.65, 114.47],
[ 1.79, 163.98],
[ 2.86, 169.55]])
In [51]: out
Out[51]:
array([[ 0. , 0. ],
[ 1.85, 0.6 ],
[ 0. , 0. ],
[ 2.73, 19.13],
[ 0. , 0. ],
[ 2.2 , 76.33],
[ 2.49, 85.85],
[ 0. , 0. ],
[ 2.55, 104.9 ],
[ 2.65, 114.47],
[ 0. , 0. ],
[ 1.79, 163.98],
[ 2.86, 169.55]])
答案 2 :(得分：2)
a=[[0.00, 0.00],
[1.85, 9.60],
[2.73, 19.13],
[0.30, 28.70],
[2.64, 38.25],
[2.29, 47.77],
[2.01, 57.28],
[2.61, 66.82],
[2.20, 76.33],
[2.49, 85.85],
[2.55, 104.90],
[2.65, 114.47],
[1.79, 123.98],
[2.86, 133.55]]
i=0
while i <len(a)-1:
if (a[i+1][1]-a[i][1])>15:
a.insert(i+1,[0,0])
i=i+1
i=i+1
for line in a :
print line
输出：
[0.0, 0.0]
[1.85, 9.6]
[2.73, 19.13]
[0.3, 28.7]
[2.64, 38.25]
[2.29, 47.77]
[2.01, 57.28]
[2.61, 66.82]
[2.2, 76.33]
[2.49, 85.85]
[0, 0]
[2.55, 104.9]
[2.65, 114.47]
[1.79, 123.98]
[2.86, 133.55]
答案 3 :(得分：1)
这是正确的算法：
arr = [ ... ]
result = []
result.append(arr[0])
for i in range(1, len(arr)):
if arr[i][1] - arr[i-1][1] > 15:
result.append([0.0,0.0])
result.append(arr[i])
print(result)
答案 4 :(得分：1)
可以处理多个填充槽的单个衬垫。在这里，我在OP示例中测试它，并使用一个修改后的值。
In [70]: np.insert(a, np.where(np.diff(a[:,1])>15)[0]+2,0, axis=0)
In [71]: Out[70]:
array([[ 0. , 0. ],
[ 1.85, 9.6 ],
[ 2.73, 19.13],
[ 0.3 , 28.7 ],
[ 2.64, 38.25],
[ 2.29, 140. ], # modified
[ 0. , 0. ],
[ 2.01, 57.28],
[ 2.61, 66.82],
[ 2.2 , 76.33],
[ 2.49, 85.85],
[ 2.55, 104.9 ],
[ 0. , 0. ],
[ 2.65, 114.47],
[ 1.79, 123.98],
[ 2.86, 133.55]])
使用where代替argmax（Colonel's回答）可以处理多个广告位。 +2是必需的，因为diff只有一个，我们正在插入。 ediff1d有更多选项来处理终点。
np.insert有各种填充策略。在这种情况下，它可能会执行与Divakar's回答类似的操作 - 创建out，并将值复制到正确的广告位。
另一个答案使用np.abs()。这可能是必需的，但在我的示例中，在0回退到140之后会添加另一个57行。
答案 5 :(得分：0)
如果numpy没有使用本地方法来做这类事情，我会感到惊讶，但我认为这也会有效：
i = 1
while i < len(lst):
if abs(lst[i][1] - lst[i-1][1]) > 15:
lst[i] = [0.0, 0.0]
# uncomment to change only the second column
# lst[i][1] = 0.0
i += 1
i += 1
输出：
>>> lst
array([[ 0. , 0. ],
[ 1.85, 9.6 ],
[ 2.73, 19.13],
[ 0.3 , 28.7 ],
[ 2.64, 38.25],
[ 2.29, 47.77],
[ 2.01, 57.28],
[ 2.61, 66.82],
[ 2.2 , 76.33],
[ 2.49, 85.85],
[ 2.55, 104.9 ],
[ 2.65, 114.47],
[ 1.79, 123.98],
[ 2.86, 133.55]])
>>>
>>> i = 1
>>> while i < len(lst):
... if abs(lst[i][1] - lst[i-1][1]) > 15:
... lst[i] = [0.0, 0.0]
... i += 1
... i += 1
...
>>> lst
array([[ 0. , 0. ],
[ 1.85, 9.6 ],
[ 2.73, 19.13],
[ 0.3 , 28.7 ],
[ 2.64, 38.25],
[ 2.29, 47.77],
[ 2.01, 57.28],
[ 2.61, 66.82],
[ 2.2 , 76.33],
[ 2.49, 85.85],
[ 0. , 0. ],
[ 2.65, 114.47],
[ 1.79, 123.98],
[ 2.86, 133.55]])
>>>