如何在csv表中进行数据的行到列转置? [英] How to do row-to-column transposition of data in csv table?
问题描述
我是脚本新手.我有一个表 (Table1.txt
),我需要创建另一个表,其中 Table1 的行按列排列,反之亦然.我已经找到了针对 Perl 和 SQL 而不是 Python 的解决方案.
I'm new to scripting. I have a table (Table1.txt
) and I need to create another table that has Table1's rows arranged in columns and vice versa. I have found solutions to this problem for Perl and SQL but not for Python.
我两天前才开始学习 Python,所以就我所掌握的:
I just started learning Python two days ago, so this is as far as I got:
import csv
import sys
with open(sys.argv[1], "rt") as inputfile:
readinput = csv.reader(inputfile, delimiter=' ')
with open("output.csv", 'wt') as outputfile:
writer = csv.writer(outputfile, delimiter=" ")
for row in readinput:
values = [row[0], row[1], row[2], row[3]]
writer.writerow([values])
这只是将列复制为列.我现在想做的是将最后一行写为 writer.writecol([values])
但似乎没有这样的命令,我还没有找到另一种写法行作为列.
This just reproduces the columns as columns. What I would have liked to do now is to write the last line as writer.writecol([values])
but it seems that there is no command like that and I haven't found another way of writing rows as columns.
推荐答案
转置迭代序列的一般解决方案是:zip(*original_list)
The solution in general to transpose a sequence of iterables is: zip(*original_list)
样本输入:
1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
计划:
with open('in.txt') as f:
lis = [x.split() for x in f]
for x in zip(*lis):
for y in x:
print(y+' ', end='')
print('
')
输出:
1 6 11
2 7 12
3 8 13
4 9 14
5 10 15
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