如何使用python中的for循环计算列表中四个相邻元素的数量？

Question

我有一个字母列表，我需要计算字母J出现为JJJJ（4 Js）的次数。 必须使用for循环（而不是.count）来完成。 我写了一些代码，但出现错误，表明索引超出范围。 而且我知道代码是不正确的，因为如果有JJJJJ（5 Js），那么我的代码会将其计为2xJJJJ（4 Js），结果将是4而不是3。

我将不胜感激任何帮助或建议。 谢谢！

data = ["A","B","A","A","A","B","J","J","J","J","B","B","B","J","J","J","J","J","B","A","A","A","J","J","J","J",]
for t in range(len(data)):
    if t<len(data) and data[t] == "J":
        if t<len(data) and data[t+1] == "J":
            if t<len(data) and data[t+2] == "J":
                if data[t+3] == "J":
                        countTTT_1 += 1
print("JJJJ appears" ,countTTT_1)

Answer 1

您可以执行以下操作：

data = 'ANDJRJEJJJJARGJJJJCSEGJJJJJESFF'

def count_j(data: str, maxj: int) -> int:
    number_of_j = 0
    number_of_repetitions = 0

    for ch in data:
            if ch == 'J':
                n, number_of_j = divmod(number_of_j, maxj)

                number_of_repetitions += n
                number_of_j += 1

    n, _ = divmod(number_of_j, maxj)

    number_of_repetitions += n

    return number_of_repetitions

print(count_j(data, 4))

Answer 2

问题

在您的代码中，您多次检查t<len(data)但这并不能防止t+3大于len(data) - 1 。

只需进行很少的修改，您就可以编写：

data = ["A","B","A","A","A","B","J","J","J","J","B","B","B","J","J","J","J","J","B","A","A","A","J","J","J","J",]
countTTT_1 = 0
for t in range(len(data)):
    if t<len(data) - 1 and data[t] == "J":
        if t<len(data) - 2 and data[t+1] == "J":
            if t<len(data) - 3 and data[t+2] == "J":
                if data[t+3] == "J":
                        countTTT_1 += 1
print("JJJJ appears" ,countTTT_1)
# ('JJJJ appears', 4)

您无需勾选t ，可以直接选择正确的范围：

data = ["A","B","A","A","A","B","J","J","J","J","B","B","B","J","J","J","J","J","B","A","A","A","J","J","J","J",]
countTTT_1 = 0
for t in range(len(data) - 3):
    if data[t] == "J":
        if data[t+1] == "J":
            if data[t+2] == "J":
                if data[t+3] == "J":
                        countTTT_1 += 1
print("JJJJ appears" ,countTTT_1)
# ('JJJJ appears', 4)

整个结构可以用sum器和切片代替sum ：

>>> s = 'ABAAABJJJJBBBJJJJJBAAAJJJJ'
>>> sum(1 for i in range(len(s) - 3) if s[i:i + 4] == 'JJJJ')
4

请注意，它像原始示例一样计算重叠的子字符串。

解

如果不想计数重叠的子字符串，则可以在每次看到'J'字符时增加一个计数器。 计数器会因其他任何字符或达到“ 4”而复位。 在这种情况下， jjjj_counter会增加：

text = 'ABAAABJJJJBBBJJJJJBAAAJJJJ'

def count_jjjj(text):
  jjjj_counter = 0
  last_js = 0
  for char in text:
    if char == 'J':
      last_js += 1
      if last_js == 4:
        jjjj_counter += 1
        last_js = 0
    else:
      last_js = 0
  return jjjj_counter

print(count_jjjj(text))
# 3
print(count_jjjj('JJJ'))
# 0
print(count_jjjj('JJJJ'))
# 1
print(count_jjjj('JJJJJ'))
# 1
print(count_jjjj('JJ JJ'))
# 0