# Corrected Evolutive Kendall’s τ Coefficients for Incomplete Rankings with Ties: Application to Case of Spotify Lists

^{1}

^{2}

^{*}

## Abstract

**:**

## 1. Introduction

- 1.
- 2.
- The use of graphs associated to the series of rankings as a tool to visualize and also to help in the definition the coefficients that summarize the “behaviour” of the series of rankings.

## 2. Preliminaries

**a**,

**b**, and

**c**). Note that Equation (1) is of this form, since $n(n-1)/2$ is the maximum number of crossings between two given rankings. The same happens with the Spearman’s $\rho $ coefficient. In [16] the Spearman’s $\rho $ for two ordinal complete rankings $\mathbf{x}=({x}_{1},{x}_{2},\dots ,{x}_{n})$ and $\mathbf{y}=({y}_{1},{y}_{2},\dots ,{y}_{n})$ with ${x}_{i},{y}_{i}\in \mathbb{N}$ is defined by

## 3. Coefficients for Two Incomplete Rankings with Ties

- 1.
- A vector to define the ordinal ranking (including the description of absent elements and tied elements);
- 2.
- A matrix to indicate the relative positions of the elements of the ranking (including absent and tied elements);
- 3.
- A formula to define the coefficients for a pair of rankings by using the entries of their associate matrices defined in the previous step.

**Example**

**1.**

## 4. Main Result

**Theorem**

**1.**

- s is the number of crossings—that is, the number of pairs $\{i,j\}$—such that ${a}_{i}<{a}_{j}$ and ${b}_{i}>{b}_{j}$, or ${a}_{i}>{a}_{j}$ and ${b}_{i}<{b}_{j}$.
- ${n}_{tu}$ is the number of pairs that are tied in only one ranking (from tie to untie or viceversa), that is, such that ${a}_{i}={a}_{j}$ and ${b}_{i}\ne {b}_{j}$, or ${a}_{i}\ne {a}_{j}$ and ${b}_{i}={b}_{j}$.

- ${n}_{\u2022\u2022}$ is the number of entries such that ${a}_{i}={b}_{i}=\u2022$;
- ${n}_{\u2022\ast}$ is the number of entries, such that ${a}_{i}=\u2022$ and ${b}_{i}\ne \u2022$;
- ${n}_{\ast \u2022}$ is the number of entries, such that ${a}_{i}\ne \u2022$ and ${b}_{i}=\u2022$.

**Proof of Theorem 1.**

**The Complete Case (C):**

**a**nor in

**b**.

- C.1.1.
- If ${a}_{i}<{a}_{j}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}\mathrm{and}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}{b}_{i}<{b}_{j}$, then ${A}_{ij}{B}_{ij}+{A}_{ji}{B}_{ji}=1\xb71+(-1)\xb7(-1)=2$.
- C.1.2.
- If ${a}_{i}>{a}_{j}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}\mathrm{and}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}{b}_{i}>{b}_{j}$, then ${A}_{ij}{B}_{ij}+{A}_{ji}{B}_{ji}=(-1)\xb7(-1)+1\xb71=2$.

- C.2.1.
- If ${a}_{i}<{a}_{j}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}\mathrm{and}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}{b}_{i}>{b}_{j}$, then ${A}_{ij}{B}_{ij}+{A}_{ji}{B}_{ji}=1\xb7(-1)+(-1)\xb71=-2$.
- C.2.2.
- If ${a}_{i}>{a}_{j}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}\mathrm{and}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}{b}_{i}<{b}_{j}$, then ${A}_{ij}{B}_{ij}+{A}_{ji}{B}_{ji}=(-1)\xb7\left(1\right)+1\xb7(-1)=-2$.

- C.3.1.
- If ${a}_{i}={a}_{j}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}\mathrm{and}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}{b}_{i}<{b}_{j}$ then ${A}_{ij}{B}_{ij}+{A}_{ji}{B}_{ji}=1\xb71+1\xb7(-1)=0$.
- C.3.2.
- If ${a}_{i}={a}_{j}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}\mathrm{and}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}{b}_{i}>{b}_{j}$ then ${A}_{ij}{B}_{ij}+{A}_{ji}{B}_{ji}=1\xb7(-1)+1\xb71=0$.
- C.3.3.
- If ${a}_{i}<{a}_{j}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}\mathrm{and}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}{b}_{i}={b}_{j}$ then ${A}_{ij}{B}_{ij}+{A}_{ji}{B}_{ji}=1\xb71+(-1)\xb71=0$.
- C.3.4.
- If ${a}_{i}>{a}_{j}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}\mathrm{and}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}{b}_{i}={b}_{j}$ then ${A}_{ij}{B}_{ij}+{A}_{ji}{B}_{ji}=(-1)\xb71+1\xb71=0$.

**The Incomplete Case (I):**

- I.3.1.
- If ${a}_{i}={b}_{i}=\u2022,{a}_{i}\ne \u2022\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}\mathrm{and}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}{b}_{j}\ne \u2022$, then ${A}_{ij}{B}_{ij}+{A}_{ji}{B}_{ji}=0\xb70+0\xb70=0$.
- I.3.2.
- If ${a}_{i}=\u2022,{a}_{j}\ne \u2022,{b}_{i}\ne \u2022\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}\mathrm{and}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}{b}_{j}=\u2022$, then ${A}_{ij}{B}_{ij}+{A}_{ji}{B}_{ji}=0\xb70+0\xb70=0$.

- I.5.1.
- If ${a}_{i}={a}_{j}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}\mathrm{and}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}{b}_{i}=\u2022\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}\mathrm{and}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}{b}_{j}\ne \u2022$, then ${A}_{ij}{B}_{ij}+{A}_{ji}{B}_{ji}=0\xb70+0\xb70=0$.
- I.5.2.
- If ${a}_{i}={a}_{j}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}\mathrm{and}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}{b}_{i}\ne \u2022\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}\mathrm{and}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}{b}_{j}=\u2022$, then ${A}_{ij}{B}_{ij}+{A}_{ji}{B}_{ji}=0\xb70+0\xb70=0$.
- I.5.3.
- If ${a}_{i}=\u2022\mathrm{and}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}{a}_{j}\ne \u2022\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}\mathrm{and}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}{b}_{i}={b}_{j}$, then ${A}_{ij}{B}_{ij}+{A}_{ji}{B}_{ji}=0\xb70+0\xb70=0$.
- I.5.4.
- If ${a}_{i}\ne \u2022\mathrm{and}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}{a}_{j}=\u2022\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}\mathrm{and}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}{b}_{i}={b}_{j}$, then ${A}_{ij}{B}_{ij}+{A}_{ji}{B}_{ji}=0\xb70+0\xb70=0$.

- I.6.1.
- If ${a}_{i}\ne {a}_{j}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}\mathrm{and}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}{b}_{i}={b}_{j}=\u2022$ then ${A}_{ij}{B}_{ij}+{A}_{ji}{B}_{ji}=(\pm 1)\xb70+(\pm 1)\xb70=0$.
- I.6.2.
- If ${a}_{i}={a}_{j}=\u2022\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}\mathrm{and}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}{b}_{i}\ne {b}_{j}$ then ${A}_{ij}{B}_{ij}+{A}_{ji}{B}_{ji}=0\xb7(\pm 1)+0\xb7(\pm 1)=0$.

- If
- ${a}_{i}<{a}_{j}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}\mathrm{and}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}{b}_{i}\ne \u2022,{b}_{j}=\u2022$ then ${A}_{ij}{B}_{ij}+{A}_{ji}{B}_{ji}=1\xb70+(-1)\xb70=0$.
- If
- ${a}_{i}>{a}_{j}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}\mathrm{and}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}{b}_{i}\ne \u2022,{b}_{j}=\u2022$ then ${A}_{ij}{B}_{ij}+{A}_{ji}{B}_{ji}=(-1)\xb70+1\xb70=0$.

**Example**

**2.**

**Remark**

**1.**

**Remark**

**2.**

**Remark**

**3.**

**Corollary**

**1.**

**Remark**

**4.**

**Remark**

**5.**

**a**and

**b**as

**Example**

**3.**

**Example**

**4.**

**Example**

**5.**

**Example**

**6.**

## 5. Treatment of More Than Two Complete Rankings. Known Results

#### 5.1. Kendall Distance for Complete Rankings with Penalty Parameters

**Definition**

**1.**

- Case 1: If i and j are not tied in $\mathbf{a}$, nor in $\mathbf{b}$. If they cross their positions when passing from $\mathbf{a}$ to $\mathbf{b}$ then ${\overline{K}}_{i,j}^{(p,\phantom{\rule{0.166667em}{0ex}}q)}=1$. Otherwise, ${\overline{K}}_{i,j}^{(p,\phantom{\rule{0.166667em}{0ex}}q)}=0$.
- Case 2: If i and j are tied in both $\mathbf{a}$ and $\mathbf{b}$. Then ${\overline{K}}_{i,j}^{(p,\phantom{\rule{0.166667em}{0ex}}q)}=q$.
- Case 3: If i and j are tied only in one ranking. Then ${\overline{K}}_{i,j}^{(p,\phantom{\rule{0.166667em}{0ex}}q)}=p$.

**Remark**

**6.**

**Remark**

**7.**

#### 5.2. Series of Complete Rankings with Ties

**Definition**

**2.**

**Definition**

**3.**

- Case 4.
- If there exists a maximal set of rankings ${\mathbf{a}}_{{t}_{1}},\cdots ,{\mathbf{a}}_{{t}_{k}}$ such that for each $\ell =1,\cdots ,k$ the pair $\{i,j\}$ is not tied in ${\mathbf{a}}_{{t}_{\ell}}$, but is tied in ${\mathbf{a}}_{{t}_{\ell}+1},{\mathbf{a}}_{{t}_{\ell}+2},\dots ,{\mathbf{a}}_{{t}_{\ell}+s}$, with $s\ge 1$, it is not tied in ${\mathbf{a}}_{{t}_{\ell}+s+1}$ and, moreover, $\{i,j\}$ exchange their relative positions between ${\mathbf{a}}_{{t}_{\ell}}$ and ${\mathbf{a}}_{{t}_{\ell}+s+1}$. In this case ${\overline{K}}_{i,j}^{cat}({\mathbf{a}}_{1},{\mathbf{a}}_{2},\dots ,{\mathbf{a}}_{m})=k$, where k is the number of rankings in the maximal set of rankings ${\mathbf{a}}_{{t}_{1}},\cdots ,{\mathbf{a}}_{{t}_{k}}$ verifying the aforementioned property.

**Example**

**7.**

**Definition**

**4.**

## 6. New Coefficients for Series of Incomplete Rankings with Ties

**Definition**

**5.**

**Definition**

**6.**

**Remark**

**8.**

**Example**

**8.**

**Example**

**9.**

**Example**

**10.**

## 7. Results

#### 7.1. Method to Convert Spotify Lists into Incomplete Rankings

- 1.
- Select a set of m lists $\{{\mathbf{v}}_{1},{\mathbf{v}}_{2},\dots ,{\mathbf{v}}_{m}\}$ with k entries in each ${\mathbf{v}}_{i}$.
- 2.
- Denote as n the number of different songs that appear on these m lists. We tag these songs from 1 to n, following the order they first appear, reading the lists from the first to the last one, and each list from top to bottom. Denote ${\mathbf{t}}_{i}$ the tagged version of ${\mathbf{v}}_{i}$, for $i=1,2,\dots ,m$, including all the n songs.
- 3.
- Denote ${\mathbf{r}}_{1}$ a vector with entries from 1 to n. The first k values correspond to the elements in ${\mathbf{v}}_{1}$.
- 4.
- Construct the rankings ${\mathbf{r}}_{i}$ for $2=1,\dots m$, in the following form:
- (a)
- The first k entries of ${\mathbf{r}}_{i}$ are copied from ${\mathbf{t}}_{i}$;
- (b)
- The rest of the entries form a vector ${\mathbf{s}}_{i}$ and come from the the elements that quit from ${\mathbf{t}}_{i-1}$ plus the elements that, being in ${\mathbf{s}}_{i-1}$, are not included in ${\mathbf{t}}_{i}$.

These $n-k$ elements preserve their relative order. This order is not important since these elements are not included in the Top k ranking ${\mathbf{t}}_{i}$. - 5.
- From each ${\mathbf{t}}_{i}$, we construct the corresponding incomplete ranking ${\mathbf{a}}_{i}$ given by (5).

**Example**

**11.**

#### 7.2. Comparison of Two Series of Top 200 Rankings

- 2019 Series: 18 weekly rankings ranging from 28 December 2018 to 3 May 2019.
- 2020 Series: 18 weekly rankings ranging from 27 December 2019 to 1 May 2020.

#### 7.3. Comparison of Two Series of Viral-50 Rankings

- 2019 Series: 18 weekly rankings ranging from 3 January 2019 to 2 May 2019.
- 2020 Series: 18 weekly rankings ranging from 2 January 2020 to 30 April 2020.

#### 7.4. Comparison of a Series of Top 200 and a Series of Viral 50 Rankings

#### 7.5. Comparison of the Evolution of Two Series of Incomplete Ranking with Ties

- (i)
- All the tracks in ${\mathbf{s}}_{1}$ are tied. That is ${\mathbf{a}}_{1}=\left[{\u2022}_{1,200}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}{\mathbf{1}}^{n-200}\right]$ where ${\u2022}_{1,200}$ is a row vector of 200 entries of the type •, and ${\mathbf{1}}^{n-200}$ is the row vector of all-ones, with $n-200$ entries, being n the total number of different tracks in the m rankings.
- (ii)
- For $i=2,3\dots m$, we consider that in ${\mathbf{s}}_{i}$ we have (at most) two buckets of tied elements. In one bucket we have the elements (if any) that come from ${\mathbf{t}}_{i-1}$. In the other bucket, we consider the rest of the elements of ${\mathbf{s}}_{i}$

**Example**

**12.**

## 8. Conclusions

- We have defined two new coefficients to characterize a series of incomplete rankings with ties in terms of the interactions mentioned above.
- We have presented a methodology to treat Spotify charts (both Top 200 and Viral 50) as a series of incomplete rankings. This methodology allows us to obtain conclusions about the movements in the lists and, therefore, on the activity of the users of the app.
- We have obtained an artificial series of incomplete rankings with ties based on Spotify Top 200 lists, to apply our coefficients and show the applicability of the method.
- The main theoretical result (Theorem 1) may serve to define new coefficients by giving weight to the interactions between pairs of elements when going from one ranking to the next one. The applications can be of interest in other fields (neuroscience, sports, bioinformatics, etc.).

## Author Contributions

## Funding

## Acknowledgments

## Conflicts of Interest

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**Figure 1.**Projected weighted graphs representing the pairs of elements that contribute to crossings (

**left panel**) and the pairs corresponding to the case tie to untie or viceversa (

**right panel**), occurring in Example 10.

**Figure 2.**Projected weighted graph representing the crossing after ties cases occurred in Example 10.

**Figure 3.**Graph based on crossings corresponding to the giant connected component of Top 200 2019 Series (

**left panel**, 360 nodes, 16,115 edges) and Top 200 2020 Series (

**right panel**, 374 nodes, 16,564 edges).

**Figure 4.**Graph based on crossings corresponding to the giant connected component of Viral-50 2019 Series (

**left panel**, 185 nodes, 1685 edges) and Viral-50 2020 Series (

**right panel**, 186 nodes, 1447 edges).

**Figure 5.**Graph based on crossings of the type from tie to untie or vice versa corresponding to the giant connected component of 2019 Series (

**left**, 377 nodes, 49,915 edges) and 2020 Series (

**right**, 457 nodes, 82,051 edges). We also have 97 isolated nodes in the 2019 Series and 99 in the 2020 Series.

**Figure 6.**Graph based on crossings of the type from tie to tie corresponding to the giant connected component of 2019 Series (

**left**, 382 nodes, 65,269 edges) and 2020 Series (

**right**, 462 nodes, 101,101 edges).

Type | Number of Pairs |
---|---|

C.1 | ${n}_{nc}$ |

C.2 | s |

C.3 | ${n}_{tu}$ |

C.4 | ${n}_{tt}$ |

Type | Number of Pairs $\{\mathit{i},\mathit{j}\}$ |
---|---|

I.1 | $\left(\genfrac{}{}{0pt}{}{{n}_{\u2022\u2022}}{2}\right)$ |

I.2 | ${n}_{\u2022\u2022}({n}_{\ast \u2022}+{n}_{\u2022\ast})$ |

I.3 | ${n}_{\u2022\u2022}{n}_{\ast \ast}+{n}_{\ast \u2022}{n}_{\u2022\ast}$ |

I.4 | $\sum _{i=1}^{{n}_{a}}\left(\genfrac{}{}{0pt}{}{{n}_{i\u2022}}{2}\right)+\sum _{i=1}^{{n}_{b}}\left(\genfrac{}{}{0pt}{}{{n}_{\u2022i}}{2}\right)$ |

I.5 | $\sum _{i=1}^{{n}_{a}}{n}_{i\ast}{n}_{i\u2022}+\sum _{i=1}^{{n}_{b}}{n}_{\ast i}{n}_{\u2022i}$ |

I.6 | $\left(\genfrac{}{}{0pt}{}{{n}_{\ast \u2022}}{2}\right)+\left(\genfrac{}{}{0pt}{}{{n}_{\u2022\ast}}{2}\right)-\sum _{i=1}^{{n}_{a}}\left(\genfrac{}{}{0pt}{}{{n}_{i\u2022}}{2}\right)-\sum _{i=1}^{{n}_{b}}\left(\genfrac{}{}{0pt}{}{{n}_{\u2022i}}{2}\right)$ |

I.7 | $n}_{\ast \ast}({n}_{\ast \u2022}+{n}_{\u2022\ast})-\sum _{i=1}^{{n}_{a}}{n}_{i\ast}{n}_{i\u2022}-\sum _{i=1}^{{n}_{b}}{n}_{\ast i}{n}_{\u2022i$ |

**Table 3.**Number of pairs $\{i,j\}$ that have some •, corresponding to Example 2. Note that the sum of all the types is, by definition in (11) , ${N}_{inc}$.

Type | Number of Pairs $\{\mathit{i},\mathit{j}\}$ |
---|---|

I.1 | $\left(\genfrac{}{}{0pt}{}{{n}_{\u2022\u2022}}{2}\right)=1$ |

I.2 | ${n}_{\u2022\u2022}({n}_{\ast \u2022}+{n}_{\u2022\ast})=8$ |

I.3 | ${n}_{\u2022\u2022}{n}_{\ast \ast}+{n}_{\ast \u2022}{n}_{\u2022\ast}=11$ |

I.4 | $\sum _{i=1}^{{n}_{a}}\left(\genfrac{}{}{0pt}{}{{n}_{i\u2022}}{2}\right)+\sum _{i=1}^{{n}_{b}}\left(\genfrac{}{}{0pt}{}{{n}_{\u2022i}}{2}\right)=1$ |

I.5 | $\sum _{i=1}^{{n}_{a}}{n}_{i\ast}{n}_{i\u2022}+\sum _{i=1}^{{n}_{b}}{n}_{\ast i}{n}_{\u2022i}=2$ |

I.6 | $\left(\genfrac{}{}{0pt}{}{{n}_{\ast \u2022}}{2}\right)+\left(\genfrac{}{}{0pt}{}{{n}_{\u2022\ast}}{2}\right)-\sum _{i=1}^{{n}_{a}}\left(\genfrac{}{}{0pt}{}{{n}_{i\u2022}}{2}\right)-\sum _{i=1}^{{n}_{b}}\left(\genfrac{}{}{0pt}{}{{n}_{\u2022i}}{2}\right)=2$ |

I.7 | ${n}_{\ast \ast}({n}_{\ast \u2022}+{n}_{\u2022\ast})-\sum _{i=1}^{{n}_{a}}{n}_{i\ast}{n}_{i\u2022}-\sum _{i=1}^{{n}_{b}}{n}_{\ast i}{n}_{\u2022i}=14$ |

${\mathbf{a}}_{1}\to {\mathbf{a}}_{2}$ | ${\mathbf{a}}_{2}\to {\mathbf{a}}_{3}$ | ${\mathbf{a}}_{3}\to {\mathbf{a}}_{4}$ | ${\mathbf{a}}_{4}\to {\mathbf{a}}_{5}$ | ${\mathbf{a}}_{5}\to {\mathbf{a}}_{6}$ | |
---|---|---|---|---|---|

${n}_{\u2022\u2022}$ | 1 | 0 | 0 | 0 | 0 |

${n}_{\u2022\ast}$ | 1 | 1 | 0 | 1 | 0 |

${n}_{\ast \u2022}$ | 0 | 0 | 1 | 0 | 3 |

${n}_{\ast \ast}$ | 6 | 7 | 7 | 7 | 5 |

s | 9 | 12 | 8 | 9 | 4 |

${n}_{tu}$ | 2 | 0 | 2 | 1 | 1 |

${n}_{tt}$ | 0 | 0 | 0 | 0 | 0 |

${N}_{inc}$ | 13 | 7 | 7 | 7 | 18 |

$\overline{n}$ | 6 | 7 | 7 | 7 | 5 |

${\widehat{\tau}}_{ev}^{\u2022}$ | −0.3333 | −0.1429 | 0.1429 | 0.0952 | 0.1000 |

${\widehat{NS}}^{\u2022}$ | 0.6667 | 0.5714 | 0.4268 | 0.4524 | 0.4500 |

${\tau}_{ev}^{\u2022}$ | −0.1786 | −0.1071 | 0.1071 | 0.0714 | 0.0357 |

$N{S}^{\u2022}$ | 0.5893 | 0.5536 | 0.4464 | 0.4643 | 0.4821 |

${\tau}_{x}$ | −0.1786 | −0.1071 | 0.1071 | 0.0714 | 0.0357 |

${\widehat{\tau}}_{x}$ | −0.3333 | −0.1429 | 0.1429 | 0.0952 | 0.1000 |

2019 Series | 2020 Series | |
---|---|---|

n | 474 | 556 |

${N}_{inc}$ | $1.6\times {10}^{6}$ | $2.4\times {10}^{6}$ |

$<{\overline{n}}_{i,i+1}>$ | 182 | 175 |

${\tau}_{ev}^{\u2022}$ | 0.1256 | 0.0836 |

$N{S}^{\u2022}$ | 0.4372 | 0.4582 |

${\widehat{\tau}}_{ev}^{\u2022}$ | 0.8540 | 0.8421 |

${\widehat{NS}}^{\u2022}$ | 0.0730 | 0.0789 |

2019 Series | 2020 Series | |
---|---|---|

n | 315 | 300 |

${N}_{inc}$ | $8.3\times {10}^{5}$ | $7.5\times {10}^{5}$ |

$<{\overline{n}}_{i,i+1}>$ | 33.6 | 35 |

${\tau}_{ev}^{\u2022}$ | 0.0067 | 0.0093 |

$N{S}^{\u2022}$ | 0.4966 | 0.4954 |

${\widehat{\tau}}_{ev}^{\u2022}$ | 0.6037 | 0.6922 |

${\widehat{NS}}^{\u2022}$ | 0.1982 | 0.1539 |

2019 Series | 2020 Series | |
---|---|---|

n | 474 | 556 |

${N}_{inc}$ | $1.4\times {10}^{6}$ | $1.7\times {10}^{6}$ |

$<{\overline{n}}_{i,i+1}>$ | 256 | 331 |

${\tau}_{ev}^{\u2022}$ | 0.2577 | 0.3108 |

$N{S}^{\u2022}$ | 0.3712 | 0.3446 |

${\widehat{\tau}}_{ev}^{\u2022}$ | 0.8848 | 0.8757 |

${\widehat{NS}}^{\u2022}$ | 0.0576 | 0.0621 |

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## Share and Cite

**MDPI and ACS Style**

Pedroche, F.; Conejero, J.A. Corrected Evolutive Kendall’s *τ* Coefficients for Incomplete Rankings with Ties: Application to Case of Spotify Lists. *Mathematics* **2020**, *8*, 1828.
https://doi.org/10.3390/math8101828

**AMA Style**

Pedroche F, Conejero JA. Corrected Evolutive Kendall’s *τ* Coefficients for Incomplete Rankings with Ties: Application to Case of Spotify Lists. *Mathematics*. 2020; 8(10):1828.
https://doi.org/10.3390/math8101828

**Chicago/Turabian Style**

Pedroche, Francisco, and J. Alberto Conejero. 2020. "Corrected Evolutive Kendall’s *τ* Coefficients for Incomplete Rankings with Ties: Application to Case of Spotify Lists" *Mathematics* 8, no. 10: 1828.
https://doi.org/10.3390/math8101828