将列表中的两个值相减

Question

my_list = [['chr1', 65419, 65433], ['chr1', 65520, 65573], ['chr1', 69037, 71585], ['chr1', 69055, 70108], ['chr1', 137621, 139379],['chr2', 65419, 65433], ['chr2', 65520, 65573], ['chr2', 69037, 71585], ['chr3', 69055, 70108]]

在列表中，将包含字符串“ chr1”，“ chr2”，“ chr3”。我想减去每个字符串的索引2-1的值，并获得'chr1'，'chr2'，'chr3']的总值

示例在前两个字符串中（65433-65419）应该减去，然后在（65573-65520）中加上，因为两个都包含'chr1'。所有列表均应如此，最终结果应如下'chr1'总计= x_value，'chr2'总计= y_value，'chr3'总计= x_value

我是一种新的python。有人可以为此建议一个代码。

my_list = [['chr1'，65419，65433]，['chr1'，65520，65573]，['chr1'，69037，71585]，['chr1'，69055，70108]，['chr1'， 137621，139379]，['chr2'，65419，65433]，['chr2'，65520，65573]，['chr2'，69037，...

Answer 1

对于大列表，您可以执行（没有硬编码的字符串匹配：）>

from itertools import groupby

my_list = [['chr1', 65419, 65433], ['chr1', 65520, 65573], ['chr1', 69037, 71585], ['chr1', 69055, 70108], ['chr1', 137621, 139379],['chr2', 65419, 65433], ['chr2', 65520, 65573], ['chr2', 69037, 71585], ['chr3', 69055, 70108]]

f = lambda x: x[0]
for k, g in groupby(sorted(my_list, key=f), key=f):
    print(k, sum(x[2] - x[1] for x in g))

# chr1 5426
# chr2 2615
# chr3 1053

Answer 2

from collections import defaultdict
my_list = [['chr1', 65419, 65433], ['chr1', 65520, 65573], ['chr1', 69037, 71585], ['chr1', 69055, 70108], ['chr1', 137621, 139379],['chr2', 65419, 65433], ['chr2', 65520, 65573], ['chr2', 69037, 71585], ['chr3', 69055, 70108]]
result = defaultdict(int)
temp = [{i[0]:i[2]-i[1]} for i in my_list]
for di in temp:
    result [ list(di.keys())[0] ] += list(di.values())[0]
for i,v in result.items():
    print(f"{i} total = {v}")

Answer 3

my_list = [['chr1', 65419, 65433], ['chr1', 65520, 65573], ['chr1', 69037, 71585], ['chr1', 69055, 70108], ['chr1', 137621, 139379],['chr2', 65419, 65433], ['chr2', 65520, 65573], ['chr2', 69037, 71585], ['chr3', 69055, 70108]]
mylist1=list()
mylist2=list()
mylist3=list()

for i in my_list:
    if i[0]=='chr1':
        mylist1.append(i[2]-i[1])
    elif i[0]=='chr2':
        mylist2.append(i[2]-i[1])
    elif i[0]=='chr3':
        mylist3.append(i[2]-i[1])

print("chr1:",sum(mylist1))
print("chr2:",sum(mylist2))
print("chr3:",sum(mylist3))

Answer 4

您可以使用字典来保存键-值对，并在有更多相同键的列表时更新它们。