例如，我在 python 中有一個(gè)列表：mylist = [1,1,1,1,1,1,1,1,1,1,1, 0,0,1,1,1,1,0,0,0,0,0, 1,1,1,1,1,1,1,1,0,0,0,0,0,0]我的目標(biāo)是找到一行中有五個(gè)或更多零的位置，然后列出發(fā)生這種情況的位置的索引，例如，輸出將是：[17,21][30,35]這是我在此處提出的其他問題中嘗試/看到的內(nèi)容：def zero_runs(a): # Create an array that is 1 where a is 0, and pad each end with an extra 0. iszero = np.concatenate(([0], np.equal(a, 0).view(np.int8), [0])) absdiff = np.abs(np.diff(iszero)) # Runs start and end where absdiff is 1. ranges = np.where(absdiff == 1)[0].reshape(-1, 2) return ranges runs = zero_runs(list)這給出了輸出：[0,10][11,12]...這基本上只是列出所有重復(fù)項(xiàng)的索引，我將如何將這些數(shù)據(jù)分成我需要的數(shù)據(jù)
3 回答
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TA貢獻(xiàn)1775條經(jīng)驗(yàn) 獲得超8個(gè)贊
您可以使用itertools.groupby，它將識(shí)別列表中的連續(xù)組：
from itertools import groupby
lst = [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0]
groups = [(k, sum(1 for _ in g)) for k, g in groupby(lst)]
cursor = 0
result = []
for k, l in groups:
if not k and l >= 5:
result.append([cursor, cursor + l - 1])
cursor += l
print(result)
輸出
[[17, 21], [30, 35]]
慕蓋茨4494581
TA貢獻(xiàn)1850條經(jīng)驗(yàn) 獲得超11個(gè)贊
使用itertools.groupbyand 的另一種方式enumerate。
首先找到零點(diǎn)和索引：
from operator import itemgetter
from itertools import groupby
zerosList = [
list(map(itemgetter(0), g))
for i, g in groupby(enumerate(mylist), key=itemgetter(1))
if not i
]
print(zerosList)
#[[11, 12], [17, 18, 19, 20, 21], [30, 31, 32, 33, 34, 35]]
現(xiàn)在只需過濾zerosList：
runs = [[x[0], x[-1]] for x in zerosList if len(x) >= 5]
print(runs)
#[[17, 21], [30, 35]]
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