Sequential pattern mining in databases with temporal uncertainty

Abstract

Temporally uncertain data widely exist in many real-world applications. Temporal uncertainty can be caused by various reasons such as conflicting or missing event timestamps, network latency, granularity mismatch, synchronization problems, device precision limitations, data aggregation. In this paper, we propose an efficient algorithm to mine sequential patterns from data with temporal uncertainty. We propose an uncertain model in which timestamps are modeled by random variables and then design a new approach to manage temporal uncertainty. We integrate it into the pattern-growth sequential pattern mining algorithm to discover probabilistic frequent sequential patterns. Extensive experiments on both synthetic and real datasets prove that the proposed algorithm is both efficient and scalable.

Appendix 1: Deriving the geographic approach

Suppose $X \sim U(x^-, x^+)$ and $Y \sim U(y^-, y^+)$ are two uniformly distributed uncertain timestamps, then the joint 2-D distribution of X and Y is:

$$\begin{aligned} f(x,y)= {\left\{ \begin{array}{ll} \frac{1}{(x^+-x^-)(y^+-y^-)}&{}\quad \text { if }\;\; x \in [x^-, x^+], y \in [y^-, y^+] \\ 0 &{}\quad \text {otherwise} \end{array}\right. } \end{aligned}$$

(28)

First of all, we define two functions $V_x(x)$ and $V_y(y)$ which restrict the possible values of x and y.

$$\begin{aligned} V_x(x)= & {} {\left\{ \begin{array}{ll} x^-,&{}\quad x< x^-\\ x,&{}\quad x^- \le x \le x^+\\ x^+,&{}\quad x > x^+ \end{array}\right. } \end{aligned}$$

(29)

$$\begin{aligned} V_y(y)= & {} {\left\{ \begin{array}{ll} y^-,&{}\quad y< y^-\\ y,&{}\quad y^- \le y \le y^+\\ y^+,&{}\quad y > y^+ \end{array}\right. } \end{aligned}$$

(30)

Given the minimal gap constraint $g_l=l$ and maximal constraint $g_h=h$ ($l < h$), we have:

$$\begin{aligned} Y \le X+h&\implies y \le x^++h \\ Y \ge X+l&\implies y \ge x^-+l \end{aligned}$$

Since $y \in [y^-, y^+]$, we also have $V_y(x^-+l) \le y \le V_y(x^++h)$. Let $P(\left\langle XY\right\rangle )$ be the probability that X and Y satisfy gap constraints, then $P(\left\langle XY\right\rangle )$ can be computed by:

$$\begin{aligned} P(\left\langle XY\right\rangle )&=\int \int _{l \le Y-X \le h}{f(x,y)}{\mathrm {d}}x{\mathrm {d}}y\nonumber \\&= \int _{V_y(x^-+l)}^{V_y(x^++h)} {\mathrm {d}}y\int _{V_x(y-h)}^{V_x(y-l)} \frac{1}{(x^+-x^-)(y^+-y^-)}{\mathrm {d}}x\nonumber \\&= \frac{1}{S}\int _{V_y(x^-+l)}^{V_y(x^++h)}[{V_x(y-l) - V_x(y-h)]} {\mathrm {d}}y \end{aligned}$$

(31)

where $S={(x^+-x^-)(y^+-y^-)}$ is a constant. Let $f(y)=V_x(y-h) - V_x(y-l)$. Notice that if $V_x(y-h)=x^+$, it implies that $V_x(y-l)=x^+$ because $y-l > y-h$, and then the integration in Eq. (31) equals to 0; similarly, $V_x(y-l)=x^-$ implies $V_x(y-h)=x^-$, which leads to $P(\left\langle XY\right\rangle )=0$. Therefore, we only need to consider the remaining four cases:

$$\begin{aligned} V_x(y-l)&=x^+, V_x(y-h)=y-h&\implies&\max (x^++l, x^-+h) \le y \le x^++h \\ V_x(y-l)&=x^+, V_x(y-h)=x^-&\implies&x^++l< y< x^-+h\\ V_x(y-l)&=y-l, V_x(y-h)=y-h&\implies&x^-+h< y < x^++l\\ V_x(y-l)&=y-l, V_x(y-h)=x^-&\implies&x^-+l \le y \le \min (x^+l,x^-+h) \end{aligned}$$

Since we also have $y^-\le y \le y^+$, the value of f(y) depends on different ranges of y:

$$\begin{aligned} f(y)={\left\{ \begin{array}{ll} x^++h-y&{} \quad \max \big (V_y(x^++l),V_y(x^-+h)\big ) \le y \le V_y(x^++h)\\ x^+-x^-&{}\quad V_y(x^++l)<y< V_y(x^-+h)\\ h-l&{}\quad V_y(x^-+h)<y < V_y(x^++l)\\ y-l-x^-&{}\quad V_y(x^-+l) \le y \le \min \big (V_y(x^++l),V_y(x^-+h)\big )\\ \end{array}\right. } \end{aligned}$$

(32)

Here we first set:

$$\begin{aligned} a_1&=V_y(x^-+l),&a_2&=V_y(x^-+h)\\ a_3&=V_y(x^++l),&a_4&=V_y( x^++h) \end{aligned}$$

Then the range $[a_1, a_4]$ is divided into disjoint sub-partitions by two points $a_2, a_3$. In order to sort the values of $a_1,a_2,a_3 \text { and }a_4$, we set:

$$\begin{aligned} b_1&=a_1,&b_2&=\min (a_2,a_3)\\ b_3&=\max (a_2,a_3),&a_4&=b_4 \end{aligned}$$

so that we have $b_1 \le b_2 \le b_3 \le b_4$. According to the law of total probability, $P(\left\langle XY\right\rangle )$ can be computed by Eq. (33), since $[b_1,b_4]=[b_1,b_2]\cup [b_2,b_3]\cup [b_3,b_4]$.

$$\begin{aligned} P(\left\langle XY\right\rangle )&= \sum _{k=1}^3{P(\left\langle XY\right\rangle |y \in [b_k,b_{k+1}])*P(y \in [b_k, b_{k+1}])}\nonumber \\&=P(\left\langle XY_k\right\rangle )*P(Y=Y_k) \end{aligned}$$

(33)

where $Y_k \sim U(b_k,b_{k+1})$ is a uniform random variable and $P(Y=Y_k)$ can be computed as:

$$\begin{aligned} P(Y=Y_k)= \int _{b_k}^{b_{k+1}}{\frac{1}{y^+-y^-}{\mathrm {d}}y} = \frac{b_{k+1} - b_k}{y^+-y^-} \end{aligned}$$

(34)

Next, we prove the correctness of Eq. (18) in three cases.

• If $Y_k \sim U[b_1,b_2]$. Now $f(y)=y-l-x^-$. According to Eq. (31), we can compute $P(\left\langle XY_k\right\rangle )$ as:

  $$\begin{aligned} P(\left\langle XY_k\right\rangle )&= \frac{1}{S_k} \int _{b_1}^{b_2}{y-l-x^-} {\mathrm {d}}y\nonumber \\&=\frac{1}{2S_k}(y-l-x^-)^2|_{b_1}^{b_2} =\frac{b_1+b_2-2l-2x^-}{2(x^+-x^-)} \end{aligned}$$

  (35)

  Meanwhile, when $y \in [b_1,b_2]$, we know that $y-l \le x^+$ and $y-h \le x^-$. According to Eq. (19), we can compute $L_1=b_1-l-x^-$ and $L_2=b_2-l-x^-$. By substituting the values of $L_1$ and $L_2$ to Eq. (18), we can compute $P(\left\langle XY\right\rangle )$ by the geographic approach in Eq. (36), which is consistent with Eq. (35).

  $$\begin{aligned} P(\left\langle XY_k\right\rangle )=\frac{L_{k+1}+L_k}{2(x^+-x^-)}=\frac{b_1+b_2-2l-2x^-}{2(x^+-x^-)} \end{aligned}$$

  (36)

• If $Y_k \sim U[b_2,b_3]$. We consider two sub-cases here.

  (a) if $V_y(x^++l) < V_y(x^-+h)$. Referring to Eq. (31), we can compute $P(\left\langle XY_k\right\rangle )$ as follows:

  $$\begin{aligned} P(\left\langle XY_k\right\rangle )&= \frac{1}{S_k} \int _{V_y(x^++l) }^{V_y(x^-+h)}{x^+-x^-} {\mathrm {d}}y\nonumber \\&=\frac{(x^+-x^-)(V_y(x^-+h)-V_y(x^++l))}{S_k}=1 \end{aligned}$$

  (37)

  Meanwhile, when $V_y(x^++l) \le y \le V_y(x^-+h)$, we have $L_2=x^+-x^-$ and $L_3=x^+-x^-$ by Eq. (19). Then, $P(\left\langle XY_k\right\rangle )$ can be computed by the geographic function in Eq. (38), which is consistent with Eq. (37).

  $$\begin{aligned} P(\left\langle XY_k\right\rangle )=\frac{L_{k+1}+L_k}{2(x^+-x^-)}=1 \end{aligned}$$

  (38)

  (b) if $V_y(x^-+h) < V_y(x^++l)$. We first have:

  $$\begin{aligned} P(\left\langle XY_k\right\rangle )&= \frac{1}{S_k} \int _{V_y(x^-+h) }^{V_y(x^++l)}{(h-l)} {\mathrm {d}}y\nonumber \\&=\frac{(h-l)(V_y(x^++l)-V_y(x^-+h))}{S_k}=\frac{h-l}{x^+-x^-} \end{aligned}$$

  (39)

  Meanwhile, $L_2=h-l$ and $L_3=h-l$, since now we have $V_y(x^-+h) \le y \le V_y(x^++l)$. And the output of the geographic function in Eq. (40) is consistent with that of Eq. (39).

  $$\begin{aligned} P(\left\langle XY_k\right\rangle )=\frac{L_{k+1}+L_k}{2(x^+-x^-)}=\frac{h-l}{(x^+-x^-)} \end{aligned}$$

  (40)

• If $Y_k \sim U[b_3,b_4]$. First, referring to Eq. (31), we can compute $P(\left\langle XY_k\right\rangle )$ as follows:

  $$\begin{aligned} P(\left\langle XY_k\right\rangle )&= \frac{1}{S_k} \int _{b_3}^{b_4}{(x^++h-y)} {\mathrm {d}}y=\frac{1}{2S_k}(x^++h-y)^2|_{b_4}^{b_3}\nonumber \\&=\frac{2x^++2h-b_3-b_4}{2(x^+-x^-)} \end{aligned}$$

  (41)

  Meanwhile, since $\max (V(x^++l), V(x^-+h)) \le y \le V(x^++h)$, we have $L_3=x^+-b_3+h$ and $L_4=x^+-b_4+h$, and then we have Eq. (42), which is consistent with Eq. (41).

  $$\begin{aligned} P(\left\langle XY_k\right\rangle ) =\frac{L_{k+1}+L_k}{2*(x^+-x^-)}=\frac{2x^++2h-b_3-b_4}{2(x^+-x^-)} \end{aligned}$$

  (42)

Therefore, the correctness of our geographic approach in Sect. 5.1 is proved.