计算 Pandas dataframe 中分组行子集中的唯一值。 （标题我尽力了）

Question

I am trying to find the fastest time of a certain sender. In the attached picture Pandas DF you will see that I have rows of IP addresses with time and SeqNo (I know there are others but they dont matter). And basically what Im trying to do is find which IP has the fastest (so smallest number in the time column) where the SeqNo are the same. So for example with SeqNo 0 the fastest IP would be 10.10.10.7 because it has the smallest value in time, which is in unix. I need to do this over all the groups of SeqNos and find which IP has fastest, so most smallest times per group of SeqNo'.

I have tried a few different for, nested for and while loops and a few different things in pandas but im having no luck. Please help out if necessary.

Answer 1

Instead of looping, pandas lets you groupby() the SeqNo and get the minimum Time per group:

index = df.groupby('SeqNo').Time.transform('min') == df.Time
df[index]

#        Source  SeqNo  Time
# 0  10.10.10.8      0     0
# 2  10.10.10.2      1     2
# 5  10.10.10.8      2     5

Based on this sample data:

df = pd.DataFrame({'Source':[f'10.10.10.{i}' for i in np.random.randint(1,9,10)],'SeqNo':[0]*2+[1]*3+[2]*5,'Time':range(10)})

#        Source  SeqNo  Time
# 0  10.10.10.8      0     0
# 1  10.10.10.6      0     1
# 2  10.10.10.2      1     2
# 3  10.10.10.4      1     3
# 4  10.10.10.2      1     4
# 5  10.10.10.8      2     5
# 6  10.10.10.7      2     6
# 7  10.10.10.7      2     7
# 8  10.10.10.3      2     8
# 9  10.10.10.8      2     9

Answer 2

You could sort the dataframe by SeqNo and Time column first. After grouping by SeqNo , use head(1) or first() to choose the first item in each group.

df = df.sort_values(['SeqNo', 'Time'], ascending=[True, True]).groupby('SeqNo').head(1)