我需要匹配两个非常大的Numpy数组(一个是20000行,另一个大约100000行),我试图构建一个脚本来有效地完成它。简单的循环遍历数组非常慢,有人可以提出更好的方法吗?这是我想要做的:数组datesSecondDict
和数组pwfs2Dates
包含日期时间值,我需要从数组pwfs2Dates
(较小的数组)中获取每个日期时间值,并查看数组中是否有类似的日期时间值(加上减去5分钟) datesSecondDict
(可能有1个以上)。如果有一个(或多个)I使用数组valsSecondDict
(它只是数字值为datesSecondDict
的数组)的值(其中一个值)填充新数组(与数组pwfs2Dates
的大小相同)。下面是@unutbu和@joaquin一个解决方案,为我工作(谢谢你们!):Numpy数组条件匹配
import time
import datetime as dt
import numpy as np
def combineArs(dict1, dict2):
"""Combine data from 2 dictionaries into a list.
dict1 contains primary data (e.g. seeing parameter).
The function compares each timestamp in dict1 to dict2
to see if there is a matching timestamp record(s)
in dict2 (plus/minus 5 minutes).
==If yes: a list called data gets appended with the
corresponding parameter value from dict2.
(Note that if there are more than 1 record matching,
the first occuring value gets appended to the list).
==If no: a list called data gets appended with 0."""
# Specify the keys to use
pwfs2Key = 'pwfs2:dc:seeing'
dimmKey = 'ws:seeFwhm'
# Create an iterator for primary dict
datesPrimDictIter = iter(dict1[pwfs2Key]['datetimes'])
# Take the first timestamp value in primary dict
nextDatePrimDict = next(datesPrimDictIter)
# Split the second dictionary into lists
datesSecondDict = dict2[dimmKey]['datetime']
valsSecondDict = dict2[dimmKey]['values']
# Define time window
fiveMins = dt.timedelta(minutes = 5)
data = []
#st = time.time()
for i, nextDateSecondDict in enumerate(datesSecondDict):
try:
while nextDatePrimDict < nextDateSecondDict - fiveMins:
# If there is no match: append zero and move on
data.append(0)
nextDatePrimDict = next(datesPrimDictIter)
while nextDatePrimDict < nextDateSecondDict + fiveMins:
# If there is a match: append the value of second dict
data.append(valsSecondDict[i])
nextDatePrimDict = next(datesPrimDictIter)
except StopIteration:
break
data = np.array(data)
#st = time.time() - st
return data
感谢, 艾娜。
感谢这么多,它完全成功了! – Aina 2011-12-20 20:48:33