+ 2

How delete all num 2?

# i want output -[1,5,7,6,7,3,8] words = [1,2,5,7,2,6,7,3,2,2,2,2,2,8]

24th May 2020, 9:09 AM
Bruklin
Bruklin - avatar
17 Answers
+ 3
# i want output -[1,5,7,6,7,3,8] words = [1,2,5,7,2,6,7,3,2,2,2,2,2,8] output = [] for i in words: if i !=2: output.append(i) print(output)
24th May 2020, 9:16 AM
Rik Wittkopp
Rik Wittkopp - avatar
+ 8
1#your list with duplicate values 2duplicates=[1,2,5,7,2,6,7,3,2,2,2,2,2,8] 3 4#print unique duplicates list 5print(list(set(duplicates))) 6 7#A list with small numbers 8smallNumbers=[2] 9 10#print the unique duplicates list without small numbers 11list(set(duplicates)-set(smallNumbers))
26th May 2020, 6:00 AM
narayanaprasad
narayanaprasad - avatar
+ 7
you may try above ☝️program
26th May 2020, 6:00 AM
narayanaprasad
narayanaprasad - avatar
+ 6
You can apply a filter on the list: print(list(filter(lambda x: x!=2, words))) or use a list comprehension: print([x for x in words if x!=2])
24th May 2020, 9:16 AM
Kuba Siekierzyński
Kuba Siekierzyński - avatar
+ 3
𝐊𝐢𝐢𝐛𝐨 𝐆𝐡𝐚𝐲𝐚𝐥 Try it this way - there are 11 2s in the words list initially, right? The iterator goes through the list and *removes* the first met instance of the argument. And then the index skipping happens. So out of 11, it will only match i==2 six times. So only 6 2s will be removed. And since .remove delete the *first met* instances, it will remove the first six 2s from the list.
24th May 2020, 10:22 AM
Kuba Siekierzyński
Kuba Siekierzyński - avatar
+ 2
𝐊𝐢𝐢𝐛𝐨 𝐆𝐡𝐚𝐲𝐚𝐥 As you wrote it .remove(2) is applied on the whole words list each time the iterated element is equal to 2. So the first time it happens is when the iterated index is equal to 1 (words[1] is the first time the condition is True). The element with index 1 gets removed and the iteration goes on. BUT... it already skips to words[2] which is now 7. And 5 gets skipped. That doesn't hurt our iterator now, but is fatal for the ending list elements, where there are consecutive 2s. The same rule applies after the first 2 is reached - it gets removed and the iterator *skips the next element* which... happens to be 2, too. So you have at least one case when 2 will not be removed. As there are similar consecutive cases, there will be more 2s left out in the end. Try removing the conditional part and see what happens.
24th May 2020, 9:33 AM
Kuba Siekierzyński
Kuba Siekierzyński - avatar
+ 2
words=[x for x in words if x!=2] thats all!!!! also you can assigne new words if do not want to change words, or make function
24th May 2020, 8:01 PM
george
george - avatar
+ 1
𝐊𝐢𝐢𝐛𝐨 𝐆𝐡𝐚𝐲𝐚𝐥 As your code iterates through the list, it is removing i, when i==2. This causes the list to get shorter, which allows the iteration to skip over sequential 2's. Hope this makes sense.
24th May 2020, 9:38 AM
Rik Wittkopp
Rik Wittkopp - avatar
+ 1
Bruklin Here is my way to remove element. https://code.sololearn.com/cAsLt04fSp6f/?ref=app Other options are doing copy or creating new list. My code is operating on the same list. Let me know if you have any questions.
24th May 2020, 10:32 AM
$¢𝐎₹𝔭!𝐨𝓝
$¢𝐎₹𝔭!𝐨𝓝 - avatar
25th May 2020, 4:16 PM
Rohit Salve
+ 1
You can also try this: words = [1,2,5,7,2,6,7,3,2,2,2,2,2,8] for i in range(words.count(2)):words.remove(2) print(words)
25th May 2020, 5:01 PM
Thameem Jabir KJ
Thameem Jabir KJ - avatar
+ 1
Every each code is good ☝☝👍👍💖
26th May 2020, 5:32 PM
Bruklin
Bruklin - avatar
0
Rohit Salve Your code is not working. I have commented to your code. Please check and solve it.
26th May 2020, 5:31 AM
$¢𝐎₹𝔭!𝐨𝓝
$¢𝐎₹𝔭!𝐨𝓝 - avatar