如何从python中的for循环逐列将数据写入csv？

Question

我有几个 for 循环，每个循环输出多行代码我需要逐列组合它们，我将附上下面的代码以解释和显示我的请求：

myFile = open('example2.csv', 'w')

links = driver.find_elements_by_css_selector('html body div#wrapper div.container.coins div.mt-3 div.col-12.row.p-0.m-0 div.col-lg-8.col-md-8.d-flex.justify-content-center.flex-md-row.align-middle.align-items-center.justify-content-md-start.p-0.m-0 div.mr-md-3.mx-2.mb-md-0.text-3xl.font-semibold')
for linka in links:
    linka = str(linka.text)
    print(linka+'\n')
links = driver.find_elements_by_xpath('/html/body/div[2]/div[4]/div[4]/div[3]/div[1]/div/div[2]')
for linkb in links:
    linkb = str(linkb.text)
    print(linkb+'\n')

links = driver.find_elements_by_css_selector('div.coin-link-row:nth-child(3)')
for linkc in links:
    linkc = str(linkc.text)
    print(linkc+'\n')

links = driver.find_elements_by_css_selector('html body div#wrapper div.container.coins div.mt-3 div.col-12.row.p-0.m-0.mb-2.d-flex.flex-column-reverse.flex-sm-row div.col-md-9.col-lg-7.p-0 div.d-flex.flex-column.mx-3.mx-sm-5.mx-md-0 div.coin-link-row.mb-md-0')
for linkd in links:
    linkd = str(linkd.text)
    print(linkd+'\n')
myData = [["Linka", "Linkb", "Linkc","Linkd"],
          [linka,linkb,linkc,linkd]]
with myFile:
    writer = csv.writer(myFile)
    writer.writerows(myData)

另外我通常会遇到这个错误

I/O operation on closed file

我需要 CSV 文件中的输出如下所示

Linka, Linkb, Linkc, Linkd
linka(loop no 1), linkb(loop no 1), linkc(loop no 1), linkd(loop no 1)
linka(loop no 2), linkb(loop no 2), linkc(loop no 2), linkd(loop no 2)
linka(loop no 3), linkb(loop no 3), linkc(loop no 3), linkd(loop no 3)
linka(loop no 4), linkb(loop no 4), linkc(loop no 4), linkd(loop no 4)

更新以下代码：

 for one in many:
                time.sleep(1)
                print(str(one.text) + '\n')
                one.click()
                with open('example2.csv', 'w') as myFile:
                    writer = csv.writer(myFile)
                    writer.writerow(["Linka", "Linkb", "Linkc", "Linkd"])
                    linksa = driver.find_elements_by_css_selector(
                        'html body div#wrapper div.container.coins div.mt-3 div.col-12.row.p-0.m-0 div.col-lg-8.col-md-8.d-flex.justify-content-center.flex-md-row.align-middle.align-items-center.justify-content-md-start.p-0.m-0 div.mr-md-3.mx-2.mb-md-0.text-3xl.font-semibold')
                    linksb = driver.find_elements_by_xpath('/html/body/div[2]/div[4]/div[4]/div[3]/div[1]/div/div[2]')
                    linksc = driver.find_elements_by_css_selector('div.coin-link-row:nth-child(3)')
                    linksd = driver.find_elements_by_css_selector(
                        'html body div#wrapper div.container.coins div.mt-3 div.col-12.row.p-0.m-0.mb-2.d-flex.flex-column-reverse.flex-sm-row div.col-md-9.col-lg-7.p-0 div.d-flex.flex-column.mx-3.mx-sm-5.mx-md-0 div.coin-link-row.mb-md-0')

                    for linka, linkb, linkc, linkd in zip(linksa, linksb, linksc, linksd):
                        writer.writerow([linka.text, linkb.text, linkc.text, linkd.text])

谢谢Barmar，你解决了我的问题（请大家投票支持@Barmar 的答案，因为我没有足够的声誉，实际上他应得的）

Answer 1

您需要使用with open 而不仅仅是with。 docs

要使用write_rows，您需要提供一个已准备好行的可迭代对象。您可以通过准备每个链接列表、链接 b 等然后将它们放在一起来做到这一点，如果您可以假设它们的数量总是相同的话。

linkbs = []
links = driver.find_elements_by_xpath('/html/body/div[2]/div[4]/div[4]/div[3]/div[1]/div/div[2]')
for linkb in links:
    linkb = str(linkb.text)
    linkbs.append(linkb)

# then put myData together
myData = [[linkas[i], linkbs[i], linkcs[i], linkds[i]] for i in range(0, len(linkas)]
# as Barmar suggested you could instead use zip()
# zip(linkas, linkbs, linkcs, linkds)
with open('example.csv', 'w') as myFile:
    writer = csv.writer(myFile)
    writer.writerows(myData)

你需要构建每个列表，我这里只做了一个例子，并且这个方法只有在每个列表长度相同的情况下才有效，否则你需要先考虑使它们长度相同或稍后捕获 indexError .

Answer 2

您需要在一个循环中处理所有链接。您可以使用zip() 来合并所有列表。

linksa = driver.find_elements_by_css_selector('html body div#wrapper div.container.coins div.mt-3 div.col-12.row.p-0.m-0 div.col-lg-8.col-md-8.d-flex.justify-content-center.flex-md-row.align-middle.align-items-center.justify-content-md-start.p-0.m-0 div.mr-md-3.mx-2.mb-md-0.text-3xl.font-semibold')
linksb = driver.find_elements_by_xpath('/html/body/div[2]/div[4]/div[4]/div[3]/div[1]/div/div[2]')
linksc = driver.find_elements_by_css_selector('div.coin-link-row:nth-child(3)')
linksd = driver.find_elements_by_css_selector('html body div#wrapper div.container.coins div.mt-3 div.col-12.row.p-0.m-0.mb-2.d-flex.flex-column-reverse.flex-sm-row div.col-md-9.col-lg-7.p-0 div.d-flex.flex-column.mx-3.mx-sm-5.mx-md-0 div.coin-link-row.mb-md-0')

with open('example2.csv', 'w') as myFile:
    writer = csv.writer(myFile)
    writer.writerow(["Linka", "Linkb", "Linkc","Linkd"])
    for linka, linkb, linkc, linkd in zip(linksa, linksb, linksc, linksd)
        writer.writerow([linka, linkb, linkc, linkd])

对于更大的循环，您应该首先在循环之外编写标题行。然后每次循环都以追加模式打开文件。

with open('example2.csv', 'w') as myFile:
    writer = csv.writer(myFile)
    writer.writerow(["Linka", "Linkb", "Linkc","Linkd"])

for one in many:
    time.sleep(1)
    print(str(one.text) + '\n')
    one.click()

    linksa = driver.find_elements_by_css_selector('html body div#wrapper div.container.coins div.mt-3 div.col-12.row.p-0.m-0 div.col-lg-8.col-md-8.d-flex.justify-content-center.flex-md-row.align-middle.align-items-center.justify-content-md-start.p-0.m-0 div.mr-md-3.mx-2.mb-md-0.text-3xl.font-semibold')
    linksb = driver.find_elements_by_xpath('/html/body/div[2]/div[4]/div[4]/div[3]/div[1]/div/div[2]')
    linksc = driver.find_elements_by_css_selector('div.coin-link-row:nth-child(3)')
    linksd = driver.find_elements_by_css_selector('html body div#wrapper div.container.coins div.mt-3 div.col-12.row.p-0.m-0.mb-2.d-flex.flex-column-reverse.flex-sm-row div.col-md-9.col-lg-7.p-0 div.d-flex.flex-column.mx-3.mx-sm-5.mx-md-0 div.coin-link-row.mb-md-0')

    with open('example2.csv', 'a') as myFile:
        writer = csv.writer(myFile)
        for linka, linkb, linkc, linkd in zip(linksa, linksb, linksc, linksd)
            writer.writerow([linka, linkb, linkc, linkd])