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[英]How to get list of indexes of all occurences of the same value in list of lists?
[英]How to get the indexes of the same values in a list?
假設我有一個這樣的列表:
l = [1, 2, 3, 4, 5, 3]
如何獲取那些重復的 3 的索引?
首先,您需要弄清楚哪些元素被重復以及在哪里重復。 我通過在字典中索引它來做到這一點。
然后你需要提取所有重復的值。
from collections import defaultdict
l = [1, 2, 3, 4, 5, 3]
_indices = defaultdict(list)
for index, item in enumerate(l):
_indices[item].append(index)
for key, value in _indices.items():
if len(value) > 1:
# Do something when them
print(key, value)
Output:
3 [2, 5]
另一種方法是像這樣過濾掉它們:
duplicates_dict = {key: indices for key, indices in _indices.items() if len(indices) > 1}
您可以使用字典理解來獲取一個 go 中的所有重復數字及其索引:
L = [1, 2, 3, 4, 5, 3, 8, 9, 9, 8, 9]
R = { n:rep[n] for rep in [{}] for i,n in enumerate(L)
if rep.setdefault(n,[]).append(i) or len(rep[n])==2 }
print(R)
{3: [2, 5],
9: [7, 8, 10],
8: [6, 9]}
使用 for 循環的等價物是:
R = dict()
for i,n in enumerate(L):
R.setdefault(n,[]).append(i)
R = {n:rep for n,rep in R.items() if len(rep)>1}
collections 中的Counter
可用於避免不必要地創建單個項目列表:
from collections import Counter
counts = Counter(L)
R = dict()
for i,n in enumerate(L):
if counts[n]>1:
R.setdefault(n,[]).append(i)
find deplicates 並循環遍歷列表以找到相應的索引位置。 不是最有效的,但有效
input_list = [1,4,5,7,1,2,4]
duplicates = input_list.copy()
for x in set(duplicates):
duplicates.remove(x)
duplicates = list(set(duplicates))
dict_duplicates = {}
for d in duplicates:
l_ind = []
dict_duplicates[d] = l_ind
for i in range(len(input_list)):
if d == input_list[i]:
l_ind.append(i)
dict_duplicates
這應該這樣做
list = [1,2,3,4,5,3]
deletes = 0;
for element in list:
if element == 3:
print(list.index(element) + deletes)
deletes = +1;
list.remove(3)
我們獲取元素的索引,刪除一個以便它可以找到下一個,並將下一個索引加 1,以便它與原始列表索引匹配。 輸出:
2
5
l = [1,2,3,4,5,3]
for i in range(len(l)):
for j in range(i + 1 , len(l)):
if l[i] == l[j]:
z = f"{l[i]} have been repeated in {i} , {j}"
print(z)
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