问题
I have a Dataframe Like:
Time Frq_1 Seq_1 Frq_2 Seq_2 Frq_3 Seq_3
12:43:04 - 30,668 - 30,670 4,620 30,671
12:46:05 - 30,699 - 30,699 3,280 30,700
12:46:17 4,200 30,700 - 30,704 - 30,704
12:46:18 3,060 30,700 4,200 30,700 - 30,700
12:46:18 3,060 30,700 4,200 30,700 - 30,700
12:46:19 3,060 30,700 4,220 30,700 - 30,700
12:46:20 3,060 30,700 4,240 30,700 - 30,700
12:46:37 - 30,698 - 30,699 3,060 30,700
12:46:38 - 30,699 3,060 30,700 4,600 30,700
12:47:19 - 30,668 - 30,669 - 30,669
12:47:20 - 30,667 - 30,667 - 30,668
12:47:20 - 30,667 - 30,667 - 30,668
12:47:21 - 30,667 - 30,667 - 30,668
12:47:21 - 30,665 - 30,665 - 30,665
12:47:22 - 30,665 - 30,665 - 30,665
12:48:35 - 30,688 - 30,690 3,020 30,690
12:49:29 4,160 30,690 - 30,691 - 30,693
I want check the total dataframe and find the result with below condition:
- Sequence_ID for which Frequency is not null
- Sequence_ID for which Frequency is Max (in case of multiple Sequence_ID with non zero Frequency)
I want my result as below:
Time Sequence_ID Frequency
12:43:04 4,620 30,671
12:46:18 4,200 30,700
12:49:29 4,160 30,690
Time = correspond to row of (Sequence_ID & Frequency)
回答1:
This turned out to be quite involved. Here we go anyway:
long_df = pd.wide_to_long(df.reset_index(), stubnames=['Seq_', 'Frq_'],
suffix='\d+', i='index', j='j')
long_df['Frq_'] = pd.to_numeric(long_df.Frq_.str.replace(',','.')
.replace('-',float('nan')))
long_df.reset_index(drop=True, inplace=True)
ix = long_df.groupby('Seq_').Frq_.idxmax()
print(long_df.loc[ix[ix.notna()].values.astype(int)])
Time Seq_ Frq_
34 12:43:04 30,671 4.62
16 12:49:29 30,690 4.16
42 12:46:38 30,700 4.60
Seems like for the sequence 30,700, the highest frequency is 4.60, not 4.20
The first step is to collapse the dataframe into three rows, one for the Time, another for the sequence and for the frequency. We can use pd.wide_to_long with the stubnames ['Seq_', 'Frq_']:
long_df = pd.wide_to_long(df.reset_index(), stubnames=['Seq_', 'Frq_'],
suffix='\d+', i='index', j='j')
print(long_df)
Time Seq_ Frq_
index j
0 1 12:43:04 30,668 -
1 1 12:46:05 30,699 -
2 1 12:46:17 30,700 4,200
3 1 12:46:18 30,700 3,060
4 1 12:46:18 30,700 3,060
5 1 12:46:19 30,700 3,060
6 1 12:46:20 30,700 3,060
7 1 12:46:37 30,698 -
8 1 12:46:38 30,699 -
9 1 12:47:19 30,668 -
10 1 12:47:20 30,667 -
11 1 12:47:20 30,667 -
12 1 12:47:21 30,667 -
13 1 12:47:21 30,665 -
14 1 12:47:22 30,665 -
15 1 12:48:35 30,688 -
16 1 12:49:29 30,690 4,160
...
The next step is to cast to float the fequencies to float, to be able to find the maximum values:
long_df['Frq_'] = pd.to_numeric(long_df.Frq_.str.replace(',','.')
.replace('-',float('nan')))
print(long_df)
Time Seq_ Frq_
index j
0 1 12:43:04 30,668 NaN
1 1 12:46:05 30,699 NaN
2 1 12:46:17 30,700 4.20
3 1 12:46:18 30,700 3.06
4 1 12:46:18 30,700 3.06
5 1 12:46:19 30,700 3.06
6 1 12:46:20 30,700 3.06
7 1 12:46:37 30,698 NaN
...
Then we can groupby Seq_ and find the indices with the highest values. One could also think of using max, but this would remove the Time column.
long_df.reset_index(drop=True, inplace=True)
ix = long_df.groupby('Seq_').Frq_.idxmax()
And finally index based on the above:
print(long_df.loc[ix[ix.notna()].values.astype(int)])
Time Seq_ Frq_
34 12:43:04 30,671 4.62
16 12:49:29 30,690 4.16
42 12:46:38 30,700 4.60
来源：https://stackoverflow.com/questions/58102325/find-max-frequency-for-every-sequence-id