duplicates

问题 Here is my situation: I have 2 tables, patient and study . Each table has its own PK using autoincrement. In my case, the pat_id should be unique. It's not declared as unique at database level since it could be non unique is some uses (it's not a home made system). I found out how to configure the system to consider the pat_id as unique, but I need now to cleanup the database for duplicated patients AND relink duplicated patients in study table to remaining unique patient , before deleting

问题 i need a Pyspark solution for Pandas drop_duplicates(keep=False) . Unfortunately, the keep=False option is not available in pyspark... Pandas Example: import pandas as pd df_data = {'A': ['foo', 'foo', 'bar'], 'B': [3, 3, 5], 'C': ['one', 'two', 'three']} df = pd.DataFrame(data=df_data) df = df.drop_duplicates(subset=['A', 'B'], keep=False) print(df) Expected output: A B C 2 bar 5 three A conversion .to_pandas() and back to pyspark is not an option. Thanks! 回答1: Use window function to count

问题 I have a list as below [('generators', 'generator'), ('game', 'games'), ('generator', 'generators'), ('games', 'game'), ('challenge', 'challenges'), ('challenges', 'challenge')] Pairs ('game', 'games') and ('games', 'game') are kind of same but they are in different order. The output I am trying to achieve [('generators', 'generator'), ('games', 'game'), ('challenge', 'challenges')] How can I remove pairs as such from above list ? Any suggestions, greatly appreciated. 回答1: You can use an

问题 I have an unsorted array of integers where the value is ranging from Integer.MIN_VALUE to Integer.MAX_VALUE. There can be multiple duplicates of any integer in the array. I need to return an array with all duplicates removed and also maintain the order of elements. example: int[] input = {7,8,7,1,9,0,9,1,2,8} output should be {7,8,1,9,0,2} I know this problem can be solved using LinkedHashSet but I need a solution which doesn't involve significant buffer space. 回答1: You can use java 8 Arrays

问题 I would like to determine whether a list contains any duplicate elements, while considering permutations as equivalent. All vectors are of equal length. What is the most efficient way (shortest running time) to accomplish this? ## SAMPLE DATA a <- c(1, 2, 3) b <- c(4, 5, 6) a.same <- c(3, 1, 2) ## BOTH OF THSE LISTS SHOULD BE FLAGGED AS HAVING DUPLICATES myList1 <- list(a, b, a) myList2 <- list(a, b, a.same) # CHECK FOR DUPLICATES anyDuplicated(myList1) > 0 # TRUE anyDuplicated(myList2) > 0 #

问题 I use to do SELECT email, COUNT(email) AS occurences FROM wineries GROUP BY email HAVING (COUNT(email) > 1); to find duplicates based on their email. But now I'd need their ID to be able to define which one to remove exactly. The second constraint is: I want only the LAST INSERTED duplicates. So if there's 2 entries with test@test.com as an email and their IDs are respectively 40 and 12782 it would delete only the 12782 entry and keep the 40 one. Any ideas on how I could do this? I've been

问题 I have an array of values like [0.75, 0.0050000000000000001, 0.0050000000000000001, 0.0050000000000000001, 0.0050000000000000001, 0.0050000000000000001, 0.0040000000000000001, ...] and I need to remove the duplicates. I only want to focus on the first 3 digits after the decimal point. How do I do this? 回答1: You can use NumberFormatter to fix the minimum and maximum fraction digits and use a set to filter the duplicate elements: let array = [0.75, 0.0050000000000000001, 0.0050000000000000001,

问题 Trying to remove duplicate based on unique values on column 'new', I have even tried two methods, but the output df.shape suggests before/after have the same df shape, meaning remove duplication fails. import pandas import numpy as np import random df = pandas.DataFrame(np.random.randn(10, 4), columns=list('ABCD')) df['new'] = [1, 1, 3, 4, 5, 1, 7, 8, 1, 10] df['new2'] = [1, 1, 2, 4, 5, 3, 7, 8, 9, 5] print df.shape df.drop_duplicates('new', take_last=False) df.groupby('new').max() print df

问题 Exmaple: [empid date bookid] ---------- 1 5/6/2004 8 2 5/6/2004 8 1 5/7/2004 8 1 5/8/2004 6 3 5/8/2004 8 2 5/8/2004 7 In this table,I need to get empid 1 as output..since it has bookid 8 more than once.. thanks in advance.. 回答1: You can use: SELECT DISTINCT id FROM table GROUP BY empid, bookid HAVING COUNT(*) > 1 But it will give you duplicates. If, for example, you have 1-8,1-8,1-9,1-9 you will get 1,1 as output because empid 1 has duplicate bookid's for two distinct bookid values. You will

问题 What would be the best way of removing any duplicate characters and sets of characters separated by spaces in string? I think this example explains it better: foo = 'h k k h2 h' should become: foo = 'h k h2' # order not important Other example: foo = 's s k' becomes: foo = 's k' 回答1: ' '.join(set(foo.split())) Note that split() by default will split on all whitespace characters. (e.g. tabs, newlines, spaces) So if you want to split ONLY on a space then you have to use: ' '.join(set(foo.split(