![](/img/trans.png)
[英]Python - group by columns A + B and count row values for columns C for each unique occurrence of A + B
[英]python group by and count Columns for each line
我有一个 txt 文件,其中包含 n 行,每行有 n 列,带有一个分隔符。
File :
x|x|x|x
x|x|x|x|x|x
x|x|x|x|x|x|x|x|x|x|x
x|x|x
x|x|x|x
x|x|x
我想像下面的输出
out:
按列分组(列数相同) - 列数 - 行号
2 - 4 - line 1, line 5
1 - 6 - line 2
1 - 11 - line 3
2 - 3 - line 4,line 6
你能帮我吗? 我尝试过熊猫,但我无法成功。
当然。 你绝对不需要 Pandas; collections.defaultdict
是你的朋友。
import io
from collections import defaultdict
# Could be a `open(...)` instead, but we're using a
# StringIO to make this a self-contained program.
data = io.StringIO("""
x|x|x|x
x|x|x|x|x|x
x|x|x|x|x|x|x|x|x|x|x
x|x|x
x|x|x|x
x|x|x
""".strip())
linenos_by_count = defaultdict(set)
for lineno, line in enumerate(data, 1):
count = line.count("|") + 1 # Count delimiters, add 1
linenos_by_count[count].add(lineno)
for count, linenos in sorted(linenos_by_count.items()):
lines_desc = ", ".join(f"line {lineno}" for lineno in sorted(linenos))
print(f"{len(linenos)} - {count} - {lines_desc}")
产出
2 - 3 - line 4, line 6
2 - 4 - line 1, line 5
1 - 6 - line 2
1 - 11 - line 3
这是使用基于@AKX 方法的itertools.groupby
的替代方法:
from itertools import groupby
print('\n'.join([f'{len(G)} - {k} - '+', '.join([f'line {x[0]+1}' for x in G])
for k, g in groupby(sorted(enumerate([s.count('x')
for s in data.split('\n')
]),
key=lambda x: x[1]),
lambda x: x[1]
)
for G in [list(g)]
]))
输出:
2 - 3 - line 4, line 6
2 - 4 - line 1, line 5
1 - 6 - line 2
1 - 11 - line 3
下面是一个不那么野蛮的格式:
from itertools import groupby
counts = [s.count('x') for s in data.split('\n')]
for k, g in groupby(sorted(enumerate(counts),
key=lambda x: x[1]),
lambda x: x[1]):
G = list(g)
print(f'{len(G)} - {k} - '+', '.join([f'line {x[0]+1}' for x in G]))
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.