计算2D列表中连续3的数量

时间:2017-10-05 17:45:26

标签: python python-2.7

我作为列表导入python的数据集:

My dataset which I imported into python as a list

有没有办法计算连续3的最大数量?与第一行一样,输出应为5,因为有5个连续的3。

import csv
r = csv.reader(open('motor.csv'))
list_r = list(r)


for row in list_r:
    print
    count = 0
    for col in row:
        if col == '3' and row[row.index(col)+1] == '3':
            count+=1

print count

这是我写的代码,但我的输出似乎不正确。

4 个答案:

答案 0 :(得分:1)

考虑使用itertools.groupby将列表分成相同值的子序列。然后简单地返回子序列的最大长度。

from itertools import groupby
list_r = [
    ['3','3','3','3','3','1','3','3','5'],
    ['1','2','3','3','3','3','3','3','1','3','3','5','3'],
    ['3','2','3','3','3','3','3','3','1','3','3','5'],
]

result = [
    max(len(list(g)) for k, g in groupby(row) if k == '3')
    for row in list_r
]

assert result == [5, 6, 6]

答案 1 :(得分:0)

他们使用以下内容作为指南:

import itertools

def consecutive(group):
    first, second = itertools.tee(group)
    second.next()
    for first, second in itertools.izip(first, second):
        if second != first + 1:  return False
    return True

def iterate_submatrix(matrix, t, l):
    '''yield the horizontals and diagonals of 4x4  subsection of matrix starting at t(op), l(eft) as 4-tuples'''
    submat =  [row[l:l+4] for row in matrix[t:t+4]]
    for r in submat: yield tuple(r)  
    for c in range (0,4):     
        yield tuple(r[c] for r in submat)
    yield tuple(submat[rc][rc] for rc in range (0,4))
    yield tuple(submat[rc][3-rc] for rc in range(0,4))

for item in iterate_submatrix(test_matrix, 0,0):
     print item, consecutive(item)

答案 2 :(得分:0)

首先,row.index(col)将始终生成行中{em> first 值col的索引。这显然不是预期的。相反,我建议使用enumerate同时迭代行中的值和索引。

其次,您只跟踪当前连续3的数量,并且没有代码可以跟踪此计数值的最大值。在代码中添加另一个变量和else子句可以解决此问题。

for row in list_r:
    max_count = current_count = 0
    for index, value in enumerate(row[:-1]):
        if value == '3' and row[index+1] == '3':
            current_count += 1
        else:
            max_count = max(current_count, max_count)
            current_count = 0
    print count

答案 3 :(得分:0)

import re

data = [
    ['1', '2', '2', '3', '5', '6'],
    ['1', '2', '3', '3', '4', '5'],
    ['1', '2', '3', '3', '3', '4']
]

max = 0
for item in data:
    match = re.search(r'3+', "".join(item))

try:

    if len(str(match.group(0))) > max:
        max = len(str(match.group(0)))
except AttributeError:
    pass

print(max)