Python在列表中找到n個連續的數字

Question

我想知道如何查找列表中是否有一定數量的連續數字，例如

例如，如果我找了兩個1的，那麼：

list = [1, 1, 1, 4, 6] #original list 
list = ["true", "true", 1, 4, 6] #after my function has been through the list. 

如果我找了三個1，那麼：

list = [1, 1, 1, 4, 6] #original list 
list = ["true", "true", "true", 4, 6] #after my function has been through the list. 

我曾嘗試：

list = [1, 1, 2, 1] 

1,1,1 in list #typed into shell, returns "(1, 1, True)" 

任何幫助將不勝感激，我主要想了解正在發生什麼，以及如何檢查列表中的下一個元素與第一個x的數量相同。

Answer 1

>>> def find_repeats(L, num_repeats): 
...  idx = 0 
...  while idx < len(L): 
...   if [L[idx]]*num_repeats == L[idx:idx+num_repeats]: 
...    L[idx:idx+num_repeats] = [True]*num_repeats 
...    idx += num_repeats 
...   else: 
...    idx += 1 
...  return L 
... 
>>> L=[1,1,1,4,6] 
>>> print find_repeats(L, 2) 
[True, True, 1, 4, 6] 
>>> L=[1,1,1,4,6] 
>>> print find_repeats(L, 3) 
[True, True, True, 4, 6] 
>>> 

這裏是一個版本，可以讓你還可以指定哪些數應匹配和第一置換後停止

>>> def find_repeats(L, required_number, num_repeats, stop_after_match=False): 
...  idx = 0 
...  while idx < len(L): 
...   if [required_number]*num_repeats == L[idx:idx+num_repeats]: 
...    L[idx:idx+num_repeats] = [True]*num_repeats 
...    idx += num_repeats 
...    if stop_after_match: 
...     break 
...   else: 
...    idx += 1 
...  return L 
... 
>>> L=[1,1,1,4,6] 
>>> print find_repeats(L, 1, 2) 
[True, True, 1, 4, 6] 
>>> L=[1,1,1,4,6] 
>>> print find_repeats(L, 1, 3) 
[True, True, True, 4, 6] 
>>> L=[1,1,1,4,4,4,6] 
>>> print find_repeats(L, 1, 3) 
[True, True, True, 4, 4, 4, 6] 
>>> L=[1,1,1,4,4,4,6] 
>>> print find_repeats(L, 4, 3) 
[1, 1, 1, True, True, True, 6] 

Answer 2

指定給list是個壞主意。使用不同的名稱。

要查找連續相等值的數量最多可以使用itertools.groupby

>>> import itertools 
>>> l = [1, 1, 1, 4, 6] 
>>> max(len(list(v)) for g,v in itertools.groupby(l)) 
3 

只搜索連續的1：

>>> max(len(list(v)) for g,v in itertools.groupby(l, lambda x: x == 1) if g) 
3 

Answer 3

我不明白什麼是你想這樣做，但我準備了一個快速而不是很好的腳本，但它可以滿足你的需求。

def repeated(num, lyst): 
# the 'out' list will contain the array you are looking for 
out = [] 
# go through the list (notice that you go until "one before 
# the end" because you peek one forward) 
for k in range(len(lyst)-1): 
    if lyst[k] == lyst[k+1] == num: 
    # if the numbers are equal, add True (as a bool, but you could 
    # also pass the actual string "True", as you have it in your question) 
    out.append(True) 
    else: 
    # if they are not the same, add the number itself 
    out.append(lyst[k]) 
# check the last element: if it is true, we are done (because it was the same as the 
# last one), if not, then we add the last number to the list (because it was not the 
# same) 
if out[-1] != True: 
    out.append(lyst[-1]) 
# return the list 
return out 

這樣使用它：

print repeated(1, [1, 1, 1, 4, 6])