使用比较和cummax：

threshold = 100
df['bool'] = df['price'].ge(threshold).cummax()

请注意，它可以以相反的方式工作（尽管可能效率较低 *）：

threshold = 100
df['bool'] = df['price'].cummax().ge(threshold)

输出量：

index  price   bool
0      0   98.0  False
1      1   99.0  False
2      2  100.5   True
3      3  101.0   True
4      4   99.0   True
5      5   98.0   True

• 实际上是在大阵列上：

%%timeit
df['price'].ge(threshold).cummax()
# 193 µs ± 4.96 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
%%timeit
df['price'].cummax().ge(threshold)
# 309 µs ± 4.9 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

计时

# setting up a dummy example with 10M rows
np.random.seed(0)
df = pd.DataFrame({'price': np.random.choice([0,1], p=[0.999,0.001], size=10_000_000)})
threshold = 0.5
## comparison
%%timeit
df['bool'] = (df.loc[:, 'price'] > threshold).cumsum().fillna(0).gt(0)
# 271 ms ± 28.8 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
%%timeit
df['bool'] = df['price'].ge(threshold).cummax()
# 109 ms ± 5.74 ms per loop (mean ± std. dev. of 7 runs, 10 loops each
%%timeit
df['bool'] = np.maximum.accumulate(df['price'].to_numpy()>threshold)
# 75.8 ms ± 2.86 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)