如何筛选包含字符串和子字符串的列表以仅返回最长的字符串.(如果列表中的任何项是另一项的子字符串,则只返回较长的字符串.)
我有这个功能.有更快的方法吗?
def filterSublist(lst):
uniq = lst
for elem in lst:
uniq = [x for x in uniq if (x == elem) or (x not in elem)]
return uniq
lst = ["a", "abc", "b", "d", "xy", "xyz"]
print filterSublist(lst)
> ['abc', 'd', 'xyz']
> Function time: 0.000011
Run Code Online (Sandbox Code Playgroud) 我有一个data.table
,我需要在它上面计算一些新值并选择具有min
值的行。
tb <- data.table(g_id=c(1, 1, 1, 2, 2, 2, 3),
item_no=c(24,25,26,27,28,29,30),
time_no=c(100, 110, 120, 130, 140, 160, 160),
key="g_id")
# g_id item_no time_no
# 1: 1 24 100
# 2: 1 25 110
# 3: 1 26 120
# 4: 2 27 130
# 5: 2 28 140
# 6: 2 29 160
# 7: 3 30 160
ts <- 118
gId <- 2
tb[.(gId), list(item_no, tdiff={z=abs(time_no - ts)})]
# g_id item_no tdiff
# 1: 2 …
Run Code Online (Sandbox Code Playgroud) 我想每月绘制数据并每年显示一年的标签.这是数据:
timedates = ['2013-01-01', '2013-02-01', '2013-03-01', '2013-04-01', '2013-05-01', '2013-06-01', '2013-07-01',
'2013-08-01', '2013-09-01', '2013-10-01', '2013-11-01', '2013-12-01', '2014-01-01', '2014-02-01',
'2014-03-01', '2014-04-01', '2014-05-01', '2014-06-01', '2014-07-01', '2014-08-01', '2014-09-01',
'2014-10-01', '2014-11-01', '2014-12-01']
timedates = pd.to_datetime(timedates)
amount = [38870, 42501, 44855, 44504, 41194, 42087, 43687, 42347, 45098, 43783, 47275, 49767,
39502, 35951, 47059, 47639, 44236, 40826, 46087, 41462, 38384, 41452, 36811, 37943]
types = ['A', 'B', 'C', 'A', 'B', 'C', 'A', 'B', 'C', 'A', 'B', 'C',
'A', 'B', 'C', 'A', 'B', 'C', 'A', 'B', 'C', 'A', …
Run Code Online (Sandbox Code Playgroud) 我想将欧几里得距离设置为LSTM或RNN的损失函数。
该函数应具有什么输出:float,(batch_size)或(batch_size,时间步长)?
模型输入X_train是(n_samples,时间步长,data_dim)。Y_train具有相同的尺寸。
示例代码:
def euc_dist_keras(x, y):
return K.sqrt(K.sum(K.square(x - y), axis=-1, keepdims=True))
model = Sequential()
model.add(SimpleRNN(n_units, activation='relu', input_shape=(timesteps, data_dim), return_sequences=True))
model.add(Dense(n_output, activation='linear'))
model.compile(loss=euc_dist_keras, optimizer='adagrad')
model.fit(y_train, y_train, batch_size=512, epochs=10)
Run Code Online (Sandbox Code Playgroud)
因此,我应该在时间步长维度和/或batch_size中平均损失吗?
python ×2
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