我正在嘗試構(gòu)建一個有助于簡化研究工作的工具，并且似乎需要檢測我何時在一列中的數(shù)據(jù)中具有遞增序列，而在另一列中具有 asc/desc 序列。有沒有一種干凈的方法來檢查行中是否有序列，而不必編寫一個像https://stackoverflow.com/a/52679427/5045375這樣遍歷行的狀態(tài)機(jī)？編寫這樣的代碼片段必須檢查一列中的值是否在遞增（無間隙），而另一列中的值是否為 asc/desc（無間隙）。我完全能夠做到這一點(diǎn)，我只是想知道我的熊貓工具箱中是否有我遺漏的東西。這里有一些例子來澄清我的意圖，import pandas as pd from collections import namedtupleQUERY_SEGMENT_ID_COLUMN = 'Query Segment Id'REFERENCE_SEGMENT_ID_COLUMN = 'Reference Segment Id'def dataframe(data): columns = [QUERY_SEGMENT_ID_COLUMN, REFERENCE_SEGMENT_ID_COLUMN] return pd.DataFrame(data, columns=columns)# No sequence in either column. No resultsdata_without_pattern = [[1, 2], [7, 0], [3, 6]]# Sequence in first column, but no sequence in second column. No resultsdata_with_pseodo_pattern_query = [[1, 2], [2, 0], [3, 6]]# Sequence in second column, but no sequence in first column. No resultsdata_with_pseudo_pattern_reference = [[1, 2], [7, 3], [3, 4]]# Broken sequence in first column, sequence in second column. No resultsdata_with_pseudo_pattern_query_broken = [[1, 2], [3, 3], [7, 4]]# Sequence occurs in both columns, asc. Expect resultsdata_with_pattern_asc = [[1, 2], [2, 3], [3, 4]]# Sequence occurs in both columns, desc. Expect resultsdata_with_pattern_desc = [[1, 4], [2, 3], [3, 2]]# There is a sequence, and some noise. Expect resultsdata_with_pattern_and_noise = [[1, 0], [1, 4], [1, 2], [1, 3], [2, 3], [3, 4]]在第一個示例中，沒有任何模式，print(dataframe(data_without_pattern)) Query Segment Id Reference Segment Id0 1 21 7 02 3 6第二個示例在查詢列中有一個升序的 id 序列，但在參考列中沒有，print(dataframe(data_with_pseodo_pattern_query)) Query Segment Id Reference Segment Id0 1 21 2 02 3 6
1 回答
繁星點(diǎn)點(diǎn)滴滴
TA貢獻(xiàn)1803條經(jīng)驗(yàn) 獲得超3個贊
如果我理解正確，您的問題是島和間隙問題的變體。每個具有可接受間隙的單調(diào)（增加或減少）子序列將形成一個島。例如，給定一個系列s：
s island
-- ------
0 1
0 1
1 1
3 2 # gap > 1, form new island
4 2
2 3 # stop increasing, form new island
1 3
0 3
概括地說：只要當(dāng)前行和前一行之間的差距超出 [-1, 1] 范圍，就會形成一個新島。
將此間隙島算法應(yīng)用于Query Segment Id和Reference Segment Id：
Query Segment Id Q Island Reference Segment Id R Island Q-R Intersection
---------------- -------- -------------------- -------- ----------------
1 1 1 1 (1, 1)
2 1 2 1 (1, 1)
3 1 3 1 (1, 1)
0 2 4 1 (2, 1)
1 2 5 1 (2, 1)
2 2 6 1 (2, 1)
3 2 7 1 (2, 1)
4 2 8 1 (2, 1)
0 3 9 1 (3, 1)
您正在尋找的qand范圍現(xiàn)在是每個 的開頭和結(jié)尾的and 。最后一個警告：忽略長度為 1 的交叉點(diǎn)（如最后一個交叉點(diǎn)）。rQuery Segment IdReference Segment IdQ-R Intersection
代碼：
columns = ['Query Segment Id', 'Reference Segment Id']
df = pd.DataFrame(data_with_multiple_contiguous_sequences, columns=columns)
def get_island(col):
return (~col.diff().between(-1,1)).cumsum()
df[['Q Island', 'R Island']] = df[['Query Segment Id', 'Reference Segment Id']].apply(get_island)
result = df.groupby(['Q Island', 'R Island']) \
.agg(**{
'Q Start': ('Query Segment Id', 'first'),
'Q End': ('Query Segment Id', 'last'),
'R Start': ('Reference Segment Id', 'first'),
'R End': ('Reference Segment Id', 'last'),
'Count': ('Query Segment Id', 'count')
}) \
.replace({'Count': 1}, {'Count': np.nan}) \
.dropna()
result['Q'] = result[['Q Start', 'Q End']].apply(tuple, axis=1)
result['R'] = result[['R Start', 'R End']].apply(tuple, axis=1)
結(jié)果：
Q Start Q End R Start R End Count Q R
Q Island R Island
1 1 1 3 1 3 3 (1, 3) (1, 3)
2 1 0 4 4 8 5 (0, 4) (4, 8)
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