Python 查找重叠或完全嵌套的范围并标记它们

如果连续的

start-stop

对的

chr

相同，我希望在按最小

start

排序到最大

start

后，在连续的

start-stop

范围中找到重叠或完全嵌套的范围

输出应如下所示：

到目前为止，我已经：

import pandas as pd
df = pd.DataFrame({'region_name': ['A', 'B', 'C', 'D', 'E', 'F', 'G', 'H'], 'start' : [1913, 46430576, 52899183, 58456122, 62925929, 65313395, 65511483, 65957829], 'stop' : [90207973, 90088654, 90088654, 74708723, 84585795, 90081985, 90096995, 83611443], 'chr':[1, 1, 1, 1, 1, 1, 1, 2]}) 

df=df.sort_值（按=['chr'，'start']，升序=[True，True]）
对于范围内的i（1，len（df['region\u name']）：
如果df['critical_error'][i]==True：
持续
对于范围（0，i）内的j：
如果df['start'][i]我没有得到零件：

。。。如果连续启动-停止对的chr相同

我仍然编写了一些代码，这些代码在某些方面与给定的表相同。如果你澄清你的观点，我可能会更新你的答案。也许它已经对您有所帮助，您可以将缺失的部分放入：

df = df.sort_values(by=['chr', 'start'], ascending=[True, True])
for i in range(1,len(df['region_name'])):
    if df['critical_error'][i] == True:
        continue
    for j in range(0,i):
        if df['start'][i] <= df['stop'][j] and df['stop'][i] <= df['stop'][j] and df['chr'][i] == df['chr'][j]:
            df['overlap'][i] ='no overlap, nested with region %s' % df['region_name'][j]
            break
        elif df['start'][i] < df['stop'][j] and df['chr'][i] == df['chr'][j]:
            df['overlap'][i] = 'overlap within region ' + df['region_name'][j]
        else:
            continue

你好这不是一个python实现，但我注意到你有基因组数据，实际上是一个非常强大、高效的工具，可以做你想要的事情。然后，您可以使用一个快速脚本进行后续操作。您好，感谢迄今为止的代码。通过相同的chr，我的意思是，对于最后一行（最后一个开始-停止对），chr是2，而对的其余部分是1，因此与其余部分没有重叠或嵌套，这是您在回答中输出的）。好的，当且仅当代码也具有相同的chr字段时，代码才会计算嵌套/重叠。如果那是你想要的，那就去吧。如果您还有任何问题，请随时提问。对不起，我无法复制您的解决方案。我在
nested=（（start>df.loc[：，'start']）和（stop无效的类型比较
错误。

。我还编辑了原始问题中的

df

，以包括

region\u name

列。您必须适应更改后的数据帧，然后：开始、停止、ch=row[：3]必须更改为新布局。是的，星号属于那里。但是您似乎使用了Python2。您可以使用以下模式替换这两个表达式：

overlap\u index=[x代表zip中的x（范围（len（overlap）），如果x[1]]

import pandas as pd
import numpy as np

df = pd.DataFrame({'region_name': ['A', 'B', 'C', 'D', 'E', 'F', 'G', 'H'], 'start' : [1913, 46430576, 52899183, 58456122, 62925929, 65313395, 65511483, 65957829], 'stop' : [90207973, 90088654, 90088654, 74708723, 84585795, 90081985, 90096995, 83611443], 'chr':[1, 1, 1, 1, 1, 1, 1, 2]})

# store texts for each row in that list
overlaps_texts = []

# iterate over all rows
for i, row in df.iterrows():
    # extract entries' data
    start, stop, ch = row[1:4]

    # Check if I am completely inside (nested into something)
    # Note that this will always return indexers where each entry if True or False
    # So nested will be something like [False, False, True, ...] where True means
    # that start > start_other AND stop < stop_other (="I am nested")
    nested = ((start > df.loc[:, 'start']) & (stop < df.loc[:, 'stop']))

    # hanging out left
    overlap_1 = ((stop > df.loc[:, 'start']) &
                 (stop < df.loc[:, 'stop'])
                 )

    # starting before stop of other but ending after (hanging out right)
    overlap_2 = ((start < df.loc[:, 'stop']) & (start > df.loc[:, 'start']))

    # one of both overlaps good
    overlap = (overlap_1 | overlap_2) & ~nested

    # identical chr? I didnt get that part. That may be different for your application
    overlap &= df.loc[:, 'chr'] == ch
    nested &= df.loc[:, 'chr'] == ch

    # generate text
    text = ''

    # check if any nestings
    if np.any(nested):
        nested_indices = [*filter(lambda x: x[1], zip(range(len(nested)), nested))]
        text = "I am nested within: "
        region_names = []
        for index, _ in nested_indices:
            region_names.append(df.iloc[index,0])

        text += ", ".join(region_names)+"; "

    # check if any overlaps (obviously one can write that more DRY), since it repeats the pattern from above
    if np.any(overlap):
        overlap_indices = [*filter(lambda x: x[1], zip(range(len(overlap)), overlap))]
        text += "I overlap: "
        region_names = []
        for index, _ in overlap_indices:
            region_names.append(df.iloc[index,0])
        text += ", ".join(region_names)

    if text == '':
        text = 'I am not nested nor do I overlap something'

    overlaps_texts.append(text)

df.loc[:, 'overlap'] = overlaps_texts

print(df)

   start                       ...                                                                 overlap
0      1913                       ...                              I am not nested nor do I overlap something
1  46430576                       ...                                     I am nested within: A; I overlap: G
2  52899183                       ...                                  I am nested within: A; I overlap: B, G
3  58456122                       ...                         I am nested within: A, B, C; I overlap: E, F, G
4  62925929                       ...                         I am nested within: A, B, C; I overlap: D, F, G
5  65313395                       ...                         I am nested within: A, B, C; I overlap: D, E, G
6  65511483                       ...                         I am nested within: A; I overlap: B, C, D, E, F
7  65957829                       ...                              I am not nested nor do I overlap something