Question

我有一个像这样的小例子的列表清单：

[['chr19', '35789598', '35789629', '21', 'chr19', '35510000', '36200000'], ['chr19', '35789598', '35789629', '24', 'chr19', '35510000', '36200000'], ['chr19', '35789598', '35789629', '52', 'chr19', '35510000', '36200000'], ['chr19', '35789598', '35789629', '88', 'chr19', '35510000', '36200000'], ['chr19', '35798974', '35799005', '56', 'chr19', '35510000', '36200000'], ['chr19', '35883830', '35883861', '16', 'chr19', '35510000', '36200000'], ['chr19', '35884320', '35884351', '51', 'chr19', '35510000', '36200000']]

如您所见，每个内部列表都有7个元素。我要创建一个新的列表列表，其中没有内部列表具有相似的第一，第二和第三元素。实际上，如果有一些内部列表，其中第1个，第2个和第3个元素是相似的，那么我将仅获取第一个内部列表，然后删除其他内部列表。小示例的预期输出如下所示：

预期输出：

[['chr19', '35789598', '35789629', '21', 'chr19', '35510000', '36200000'], ['chr19', '35798974', '35799005', '56', 'chr19', '35510000', '36200000'], ['chr19', '35883830', '35883861', '16', 'chr19', '35510000', '36200000'], ['chr19', '35884320', '35884351', '51', 'chr19', '35510000', '36200000']]

这是python中的代码，不会返回我期望的值：

result = []
for i in mat:
    for j in i:
        if j == j-1:
            result.append(j)

Answer 1

我会用熊猫：

import pandas as pd
data = [['chr19', '35789598', '35789629', '21', 'chr19', '35510000', '36200000'], 
        ['chr19', '35789598', '35789629', '24', 'chr19', '35510000', '36200000'],
        ['chr19', '35789598', '35789629', '52', 'chr19', '35510000', '36200000'], 
        ['chr19', '35789598', '35789629', '88', 'chr19', '35510000', '36200000'], 
        ['chr19', '35798974', '35799005', '56', 'chr19', '35510000', '36200000'], 
        ['chr19', '35883830', '35883861', '16', 'chr19', '35510000', '36200000'], 
        ['chr19', '35884320', '35884351', '51', 'chr19', '35510000', '36200000']]
# Convert your list of list to a DataFrame
df = pd.DataFrame(data)
       0         1         2   3      4         5         6
0  chr19  35789598  35789629  21  chr19  35510000  36200000
1  chr19  35789598  35789629  24  chr19  35510000  36200000
2  chr19  35789598  35789629  52  chr19  35510000  36200000
3  chr19  35789598  35789629  88  chr19  35510000  36200000
4  chr19  35798974  35799005  56  chr19  35510000  36200000
5  chr19  35883830  35883861  16  chr19  35510000  36200000
6  chr19  35884320  35884351  51  chr19  35510000  36200000

df = df.drop_duplicates([0, 1, 2], keep='first')
       0         1         2   3      4         5         6
0  chr19  35789598  35789629  21  chr19  35510000  36200000
4  chr19  35798974  35799005  56  chr19  35510000  36200000
5  chr19  35883830  35883861  16  chr19  35510000  36200000
6  chr19  35884320  35884351  51  chr19  35510000  36200000

# If you need the data as the list of lists still output like this:
output = df.values
array([['chr19', '35789598', '35789629', '21', 'chr19', '35510000', '36200000'],
       ['chr19', '35798974', '35799005', '56', 'chr19', '35510000', '36200000'],
       ['chr19', '35883830', '35883861', '16', 'chr19', '35510000', '36200000'],
       ['chr19', '35884320', '35884351', '51', 'chr19', '35510000', '36200000']], 
       dtype=object)

# Otherwise you can continue to use the DataFrame for your analysis

总结python中的列表列表并创建一个新矩阵

1 个答案: