[英]Break list into list of lists based on condition
我有以下列表 x1、x2、x3,我想將它們分塊到下面提到的相應輸出中:
x1 = ['req', 'a', 'b', 'c', 'req', 'd', 'e', 'req', 'f']
expected_out1 = [['req', 'a', 'b', 'c'], ['req', 'd', 'e'], ['req', 'f']]
x2 = ['req', 'a', 'b', 'c', 'req', 'd', 'e', 'req', 'f', 'req']
expected_out2 = [['req', 'a', 'b', 'c'], ['req', 'd', 'e'], ['req', 'f'], ['req']]
x3 = ['req', 'a', 'b', 'c', 'req', 'd', 'e', 'req', 'req']
expected_out3 = [['req', 'a', 'b', 'c'], ['req', 'd', 'e'], ['req'], ['req']]
我編寫了以下代碼來解決這些情況:
import numpy as np
def split_basedon_condition(b):
num_arr = np.array(b)
arrays = np.split(num_arr, np.where(num_arr[:-1] == "req")[0])
return [i for i in [i.tolist() for i in arrays] if i != []]
但我得到以下結果:
split_basedon_condition(x1)
actual_out1 = [['req', 'a', 'b', 'c'], ['req', 'd', 'e'], ['req', 'f']] # expected
split_basedon_condition(x2)
actual_out2 = [['req', 'a', 'b', 'c'], ['req', 'd', 'e'], ['req', 'f', 'req']] # not expected
split_basedon_condition(x3)
actual_out3 = [['req', 'a', 'b', 'c'], ['req', 'd', 'e'], ['req', 'req']] # not expected
原因在這一行:
arrays = np.split(num_arr, np.where(num_arr[:-1] == "req")[0])
通過執行num_arr[:-1]
您正在考慮 num_arr 放棄最后一個元素,因此當"req"
是最后一個元素時這種行為。 將上面的行替換為:
arrays = np.split(num_arr, np.where(num_arr == "req")[0])
並且對於您提供的所有測試用例,它都可以正常工作。
作為旁注,如果您被允許使用除numpy
之外的外部 python 庫,您可以利用more_itertools.split_before來完成該任務。
這是更快的純 python 解決方案。
def split(arr, pred):
j = len(arr)
for i,_ in filter(lambda x: x[1] == pred, zip(range(j-1, -1, -1), reversed(arr))):
yield arr[i:j]
j = i
list(split(['req', 'a', 'b', 'c', 'req', 'd', 'e', 'req', 'req'], 'req'))
# [['req'], ['req'], ['req', 'd', 'e'], ['req', 'a', 'b', 'c']]
如果第一個總是"req"
。這是一行:
def func(l):
return list(map(lambda x: x.insert(0, "req") or x, map(list, "".join(l[1:]).split("req"))))
結果:
[['req', 'a', 'b', 'c'], ['req', 'd', 'e'], ['req', 'f']]
[['req', 'a', 'b', 'c'], ['req', 'd', 'e'], ['req', 'f'], ['req']]
[['req', 'a', 'b', 'c'], ['req', 'd', 'e'], ['req'], ['req']]
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