快速打开和关闭csv？

Question

我正在用python编写一个程序，将csv转换为列表列表。它为不同的文件做了多次，所以我把它做成了一个函数。我还没有遇到过这样的错误，但我担心这是最具python/最聪明/最快的方式，因为这些都是巨大的csv。你知道吗

import csv

searchZipCode = #there's a zip code here
zipCoords = #there's a file here

def parseFile(selected):
    with open(selected) as selectedFile:
            selectedReader = csv.reader(selectedFile, delimiter=',')
            for row in selectedReader:
                    yield row

def parseZips():
    return parseFile(zipCoords)

zips = parseZips()
for row in zips:
    if row[0] == searchZipCode:
            searchState = row[1]
            searchLat   = row[2]
            searchLong  = row[3]
            print searchState 

基本上，我想知道为什么for row必须重复两次。难道没有更优雅的解决方案吗？你知道吗

Answer 1

您可以在读取行时简单地进行比较，而不是先产生行，然后进行迭代。你知道吗

def findZip(selected, search):
    results = []
    with open(selected) as file:
        handle = csv.reader(file, delimiter=',')
        for row in handle:
            if row[0] == search
                results.append(row[1:4])
    return results

如果您希望进一步优化它，那么一旦找到匹配项，就可以跳出循环，前提是只有一个匹配项。你知道吗

def findZip(selected, search):
    with open(selected) as file:
        handle = csv.reader(file, delimiter=',')
        for row in handle:
            if row[0] == search
                return row[1:4]