如何在python中加速for循环

Question

我正在处理两组三个相同大小的大列表，其中包含UTM格式的经度、纬度和高度坐标（见下面的列表）。数组包含重叠的坐标（即经度和纬度值相等）。如果Lon中的值等于Lon2，Lat中的值等于Lat2，那么我要计算这些指标的平均高度。但是，如果它们不相等，那么经度、纬度和高度值将保持不变。我只想将重叠的数据替换为一组经纬度坐标，并计算这些坐标的平均值。你知道吗

这是我目前为止的尝试

 import numpy as np

 Lon = [450000.50,459000.50,460000,470000]
 Lat = [5800000.50,459000.50,500000,470000]
 Alt = [-1,-9,-2,1]
 Lon2 = [450000.50,459000.50,460000,470000]
 Lat2 = [5800000.50,459000.50,800000,470000]
 Alt2= [-3,-1,-20,2]

 MeanAlt = []
 appendAlt = MeanAlt.append
 LonOverlap = []
 appendLon = LonOverlap.append
 LatOverlap = []
 appendLat = LatOverlap.append

 for i, a in enumerate(Lon and Lat and Alt):
     for j, b in enumerate(Lon2 and Lat2 and Alt2):
         if Lon[i]==Lon2[j] and Lat[i]==Lat2[j]:
             MeanAltData = (Alt[i]+Alt2[j])/2
             appendAlt(MeanAltData)
             LonOverlapData = Lon[i]
             appendLat(LonOverlapData)
             LatOverlapData = Lat[i]
             appendLon(LatOverlapData)

 print(MeanAlt) # correct ans should be MeanAlt = [-2.0,-5,1.5]
 print(LonOverlap)
 print(LatOverlap)

我在一个jupyter笔记本电脑里工作，我的笔记本电脑速度很慢，所以我需要让这个代码更有效率。如果能帮上忙，我将不胜感激。谢谢：）

Answer 1

我相信您的代码可以通过两种方式得到改进：

• 首先，使用tuples而不是lists，因为迭代元组比迭代列表要generally faster。你知道吗
• 其次，您的for循环可以简化为只有一个循环，它迭代您要读取的元组的索引。当然，这个假设适用于当且仅当所有元组包含相同数量的项（即：len(Lat) == len(Lon) == len(Alt) == len(Lat2) == len(Lon2) == len(Alt2)）。你知道吗

下面是改进的代码（我冒昧地删除了import numpy语句，因为它没有在您提供的代码中使用）：

# use of tuples
Lon = (450000.50, 459000.50, 460000, 470000)
Lat = (5800000.50, 459000.50, 500000, 470000)
Alt = (-1, -9, -2, 1)
Lon2 = (40000.50, 459000.50, 460000, 470000)
Lat2 = (5800000.50, 459000.50, 800000, 470000)
Alt2 = (-3, -1, -20, 2)

MeanAlt = []
appendAlt = MeanAlt.append
LonOverlap = []
appendLon = LonOverlap.append
LatOverlap = []
appendLat = LatOverlap.append

# only one loop
for i in range(len(Lon)):
    if (Lon[i] == Lon2[i]) and (Lat[i] == Lat2[i]):
        MeanAltData = (Alt[i] + Alt2[i]) / 2
        appendAlt(MeanAltData)
        LonOverlapData = Lon[i]
        appendLat(LonOverlapData)
        LatOverlapData = Lat[i]
        appendLon(LatOverlapData)

print(MeanAlt)  # correct ans should be MeanAlt = [-2.0,-5,1.5]
print(LonOverlap)
print(LatOverlap)

我在笔记本电脑上执行了这个程序一百万次。按照我的代码，所有执行所需的时间是：1.41秒。另一方面，使用你的方法所需的时间是：4.01秒。你知道吗

Answer 2

这不是100%的功能等效，但我猜它更接近您实际需要的：

Lon = [450000.50,459000.50,460000,470000]
Lat = [5800000.50,459000.50,500000,470000]
Alt = [-1,-9,-2,1]
Lon2 = [40000.50,459000.50,460000,470000]
Lat2 = [5800000.50,459000.50,800000,470000]
Alt2= [-3,-1,-20,2]

MeanAlt = []
appendAlt = MeanAlt.append
LonOverlap = []
appendLon = LonOverlap.append
LatOverlap = []
appendLat = LatOverlap.append

ll = dict((str(la)+'/'+str(lo), al) for (la, lo, al) in zip(Lat, Lon, Alt))

for la, lo, al in zip(Lon2, Lat2, Alt2):
    al2 = ll.get(str(la)+'/'+str(lo))
    if al2:
        MeanAltData = (al+al2)/2
        appendAlt(MeanAltData)
        LonOverlapData = lo
        appendLat(LonOverlapData)
        LatOverlapData = la
        appendLon(LatOverlapData)

print(MeanAlt) # correct ans should be MeanAlt = [-2.0,-5,1.5]
print(LonOverlap)
print(LatOverlap)

或更简单：

Lon = [450000.50,459000.50,460000,470000]
Lat = [5800000.50,459000.50,500000,470000]
Alt = [-1,-9,-2,1]

Lon2 = [40000.50,459000.50,460000,470000]
Lat2 = [5800000.50,459000.50,800000,470000]
Alt2= [-3,-1,-20,2]

ll = dict((str(la)+'/'+str(lo), al) for (la, lo, al) in zip(Lat, Lon, Alt))

result = []
for la, lo, al in zip(Lon2, Lat2, Alt2):
    al2 = ll.get(str(la)+'/'+str(lo))
    if al2:
        result.append((la, lo, (al+al2)/2))

print(result)

在实践中，我会尝试从更好的结构化输入数据开始，使得转换为dict，或者至少是“zip（）”是不必要的。你知道吗

Answer 3

使用numpy将计算矢量化。对于1000000个长数组，如果输入已经是numpy.ndarray，那么执行时间应该是15-25ms微秒，如果输入是Python列表，那么执行时间应该是140ms。你知道吗

import numpy as np
def mean_alt(lon, lon2, lat, lat2, alt, alt2):
    lon = np.asarray(lon)
    lon2 = np.asarray(lon2)
    lat = np.asarray(lat)
    lat2 = np.asarray(lat2)
    alt = np.asarray(alt)
    alt2 = np.asarray(alt2)
    ind = np.where((lon == lon2) & (lat == lat2))
    mean_alt = (0.5 * (alt[ind] + alt2[ind])).tolist()
    return (lon[ind].tolist(), lat[ind].tolist(), mean_alt)