列表中的連續值（python）

Question

我正在嘗試查找以下列表是否包含超過 4 個連續值。

小時=[9,10,11,12,13,14,15]

我試過這段代碼並返回上面的列表有超過 4 個連續值，但它不適用於其他數字。

hours=[9,28,11,23,13,9,15]
y = sorted(hours)
consecutive_hours = 0

while consecutive_hours <=4:
    for hours in hours:
        if len(y) == 7 and y[6]-y[0] == 6: 
            consecutive_hours += 1
        else:
            break

if consecutive_hours > 4:
    print("The list", hours, "contains more than 4 consecutive hours.")
else:
    print("The list",hours, "does not contain more than 4 consecutive hours.")

Answer 1

嘗試這個！

hours = [9, 28, 11, 23, 13, 9, 15]
y = sorted(hours)

# Starts with 1 consecutive hour, we will subtract later
consecutive_hours = 1

# Check for the number of consecutive hours
for i in range(len(y)):
    if y[i] == y[i - 1] + 1:
        consecutive_hours += 1

if consecutive_hours == 1: consecutive_hours = 0

# Print results
if consecutive_hours > 4:
    print("The list ", hours, " contains more than 4 consecutive hours.")
else:
    print("The list ", hours, " does not contain more than 4 consecutive hours.")

Answer 2

你可以試試這樣的

hours=[9,28,11,23,13,9,15]
y = sorted(hours)
consecutive_hours = 1

for i in range(0, len(y)):
        j = i + 1
        if j >= len(y):
            break
        if y[j] == y[i] + 1:
            consecutive_hours += 1
        elif consecutive_hours > 4:
            break
        else:
            consecutive_hours = 1

if consecutive_hours > 4:
    print("The list", hours, "contains more than 4 consecutive hours.")
else:
    print("The list",hours, "does not contain more than 4 consecutive hours.")

Answer 3

以下代碼將允許您檢查列表中的下一個數字是否是for循環所在的當前數字的consecutive_hours數字，如果是，它將向 Continuous_hours 變量添加一。 如果for循環到達的數字不是前一個數字的consecutive_hours數字，它將設置 Continuous_hours 為0 。 如果在for循環遍歷整個列表之前， consecutive_hours變量大於4 ，則for循環將中斷，如果在循環到達列表末尾時該值小於或等於4 ，則循環將結束. for循環之后的以下if語句將檢查consecutive_hours值，並打印相應的語句。

hours=[9,28,11,23,13,14,15,16,17,9,15,1,2,3,4]
y = sorted(hours)
consecutive_hours = 0

for num in hours:
    if hours.index(num) == len(hours)-1:
        break
    elif hours[hours.index(num)+1] == num + 1:
        consecutive_hours += 1
        if consecutive_hours > 4:
            break
    else:
        consecutive_hours = 0

if consecutive_hours == 4:
    print("The list", hours, "contains more than 4 consecutive hours.")
else:
    print("The list",hours, "does not contain more than 4 consecutive hours.")

Answer 4

如果不進行排序，您可以使用計數器 class（來自集合）來計算將所有值與其偏移量 0、1、2 和 3 組合時獲得的匹配數：

from collections import Counter
def maxConsec(hours,span=4):
    counts = Counter(h-i for h in {*hours} for i in range(span))
    return span in counts.values()

output：

maxConsec([9,10,11,12,13,14,15]) # True
maxConsec([9,28,11,23,13,9,15])  # False