Re: [問題] numpy問題請教
已經很接近了，但因為不知道你取incremental subset是否也要包括第一個
所以我提供了比較囉嗦但彈性一點的做法
以下兩個numpy方法提供參考
https://gist.github.com/benbenbang/8e947fbd3c40c130ec99347f9c355873
------
import numpy as np
# Set an anchor
# This won't include the first element
# So you can prevent getting [5, 6] instead of [6] in a case like [5, 6, 8, 4]
# But if you like to include 5, then simply assign 0 to the idx
def method_np_diff(l, idx=1):
ary = np.array(l).reshape(-1)
# Stick with numpy diff
diff = np.diff(ary, append=ary[0])
# This will give you array([3, 4])
increasing_subset = ary[idx:][
(np.diff(ary, append=ary[0]) > 0)[idx:]
& (np.diff(ary, prepend=ary[-1]) > 0)[idx:]
]
return diff, increasing_subset
def method_np_roll(l, idx=1):
ary = np.array(l).reshape(-1)
# You can also try numpy roll
diff = np.roll(ary, -1) - ary
# This will give you array([3, 4]) as well
increasing_subset = ary[idx:][
(np.roll(ary, -1) - ary > 0)[idx:] & (ary - np.roll(ary, 1) > 0)[idx:]
]
return diff, increasing_subset
l = [5, 2, 3, 4, 6, 1]
print(
"Diff: %(diff)s | Incremental Subset: %(subset)s"
% {"diff": method_np_diff(l)[0], "subset": method_np_diff(l)[1]}
)
# 38.6 μs ± 2.67 μs per loop (mean ± std. dev. of 7 runs, 10000 loops
each)
%timeit method_np_diff(l)
# 38.1 μs ± 661 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
%timeit method_np_roll(l)
l = np.random.randn(10000000, 1)
# 115 ms ± 518 μs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
%timeit method_np_diff(l)
# 144 ms ± 293 μs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
%timeit method_np_roll(l)
基本上效能不會差太多，但一定比list comprehension或map + lambda取值好多了
歡迎回饋任何意見
※ 引述《xAyax (willy10155170)》之銘言：
: 有幾個問題想要請教一下
: 如果想要比較一個一維陣列的每元素值
: 是否大於前一個且小於後一個
: 不用for用內建函式該怎麼做？
: Ex. A=[5, 2, 3,4,6,1]
: 我想取3,4因為2<3<4, 3<4<6
: 應該用np.where嗎？
: 可是這樣condition該怎麼填 囧
: 還有另一個問題是
: 如果有個二維陣列存各個點
: 我想計算所有各點間的距離
: 公式沒問題
: 不過我要如何做到所有排列
: 一樣不用for用內建函式的話
: Ex.[[點a],[點b],[點c]]
: 我想要計算ab, bc, ac間的距離
: 可是用np.diff只能算到ab,bc而已
: 我要如何做到連ac都算
: 希望有高人能指導一下
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