搭配詞
Collocations
1890–1960
「觀其伴，知其義。」— J. R. Firth (1957)
“You shall know a word by the company it keeps.”
DHG 502 Digital Approaches in Historical Research · ← → navigate · 1–9 sections · F fullscreen
什麼是搭配詞？What is a collocation?
兩個詞傾向於在自然語言中彼此靠近出現，稱為「共現」（cooccur）。
A combination of two words that tend to occur near each other in natural language, i.e. to cooccur.
Evert, “Corpora and collocations”, p. 1214
有些詞互相「吸引」Some words attract each other
cow – milkday – nightring – bell喝 – 茶天氣 – 好
有些則不然Others do not
know – glassdoor – year喝 – 桌子
問題：如何「測量」這種吸引力，並與偶然區分開？
How do we measure attraction, and tell it apart from chance?
隨機的詞 vs. 真實的語言Random words vs. real language
虛無假說：詞語是隨機分布的Null hypothesis: words are distributed randomly
自然語言：詞語的分布是有意義的Natural language: words are distributed meaningfully
搭配詞 ＝ 相鄰出現的頻率高於隨機打亂所預期的詞對
A collocation is a pair that sits together more often than random shuffling would predict.
觀察值 OObserved frequency: how often does Y appear near the target word X?
X 目標詞：王婆Target word
視窗 每個「王婆」的左右各 3 個詞（horizon = 3）Window: 3 words to the left and right of every 王婆. The numbers show the distance.
Y 搭配詞：痛苦，在視窗內出現 2 次Collocate: 痛苦 appears twice inside a window (in this excerpt)
\(O(\text{王婆}, \text{痛苦}) =\) 2
期望值 EExpected frequency: what would we see if it were all chance?
10 顆球，其中 5 顆紅球：\(P(\text{紅}) = 5/10 = 0.5\)10 balls, 5 of them red
抽出 4 顆，預期有幾顆紅球？（這一次：2）Draw 4 balls. How many should be red?
\(E(\text{紅} \mid 4) = P(\text{紅}) \times 4 = 0.5 \times 4 = \mathbf{2}\)
單次可能是 1 或 3 顆；期望值是「平均」：抽得越多，平均越接近 2。One draw may give 1 or 3. E is the average: the more we draw, the closer we get to 2.
隨機抽取：平均會靠近期望值Random draws: the running average approaches E
平均紅球數 vs. 抽取次數running average of red balls vs. number of draws
這一次：2 顆紅球this draw
已抽 0 次；平均 –draws so far · running average
R 抽一次
T 連抽 10 次
C 清除R: draw once · T: draw 10 times · C: clear
T 連抽 10 次
C 清除R: draw once · T: draw 10 times · C: clear
期望值：王婆 ／ 痛苦Expected frequency of the pair (王婆, 痛苦) by chance
語料總詞數 Ntotal tokens
50,000
count(王婆)times 王婆 occurs
200
count(痛苦)times 痛苦 occurs
50
視窗window
3 + 3
示意數字 illustrative numbers, not counted from the novel
① 隨機抽一個詞，剛好是「痛苦」的機率chance that a random word is 痛苦
\(P(\text{痛苦}) = \frac{50}{50{,}000} = 0.001\)
② 「王婆」周圍有多少個位置？how many slots around all 王婆?
\(200 \times 3 \times 2 = 1{,}200\)
③ 期望值 ＝ 機率 × 位置數expected = probability × slots
\(E = 0.001 \times 1{,}200 = \mathbf{1.2}\)
純屬偶然，只預期出現約 1.2 次by chance alone we expect only about 1.2 co-occurrences
觀察值 vs. 期望值Observed vs. expected for (王婆, 痛苦)
期望值 Eexpected by chance
1.2
觀察值 Oobserved in the corpus (illustrative)
8
\(\dfrac{O}{E} = \dfrac{8}{1.2} \approx \textcolor{#e0570f}{\mathbf{6.7}}\)
實際共現次數是偶然預期的 6.7 倍 → 王婆與痛苦互相「吸引」They co-occur 6.7 times more often than chance predicts: the two words attract each other.
接下來：先做一個練習，再用對數把 O/E 變成互信息 MI。Next: a practice example, then the logarithm of this ratio gives mutual information (MI).
練習：自己算 E 和 O/EPractice: compute E and O / E yourself
語料總詞數 Ntotal tokens
20,000
count(喝)times 喝 occurs
100
count(茶)times 茶 occurs
40
視窗window
4 + 4
觀察值 Oobserved (喝, 茶)
12
示意數字 illustrative numbers
① 隨機抽一個詞，剛好是「茶」的機率chance that a random word is 茶
\(P(\text{茶}) = \dfrac{40}{20{,}000} = \mathbf{0.002}\)
② 「喝」周圍有多少個位置？how many slots around all 喝?
\(100 \times 4 \times 2 = \mathbf{800}\)
③ 期望值 ＝ 機率 × 位置數expected = probability × slots
\(E = 0.002 \times 800 = \mathbf{1.6}\)
④ 觀察值 ÷ 期望值observed / expected
\(\dfrac{O}{E} = \dfrac{12}{1.6} = \textcolor{#e0570f}{\mathbf{7.5}}\)
O 遠高於偶然預期 → 喝與茶互相「吸引」O is far above chance: 喝 and 茶 attract each other.
預備知識：對數 log₂Background: the base-2 logarithm
log₂ x ＝ 2 的幾次方會等於 x？log₂ x is the power you raise 2 to in order to get x
\(2^{3} = 8 \;\Rightarrow\; \log_2 8 = \mathbf{3}\)
\(2^{0} = 1 \;\Rightarrow\; \log_2 1 = \mathbf{0}\)
\(2^{-1} = \tfrac{1}{2} \;\Rightarrow\; \log_2 \tfrac{1}{2} = \mathbf{-1}\)
比值 ×2 → 分數 ＋1
比值 ÷2 → 分數 −1
比值 ＝ 1（O ＝ E）→ 分數 0ratio doubles: +1 · ratio halves: −1 · ratio 1 (O = E): 0
比值 ÷2 → 分數 −1
比值 ＝ 1（O ＝ E）→ 分數 0ratio doubles: +1 · ratio halves: −1 · ratio 1 (O = E): 0
互信息 MIMutual information (pointwise): the log of O / E
對 O/E 取以 2 為底的對數Take the base-2 logarithm of O / E
\(\mathrm{MI} = \log_2 \dfrac{O}{E}\)
王婆 ／ 痛苦：for our example pair
\(\mathrm{MI} = \log_2 \dfrac{8}{1.2} = \log_2 6.7 \approx \textcolor{#e0570f}{\mathbf{2.74}}\ \text{bits}\)
MI 只看比值，不看證據多寡：罕見詞對容易得高分。MI only sees the ratio, not how much evidence there is: rare pairs can score very high (Evert, p. 1227).
為什麼取對數？每多一倍 ＝ ＋1Why the logarithm? Every doubling adds 1 bit
PPMI：正的點互信息PPMI: positive pointwise mutual information
MI 為負：兩詞「互相排斥」，比偶然更少共現Negative MI: the words repel each other, co-occurring less than chance
PPMI：把負值一律設為 0PPMI: replace every negative value with 0
\(\mathrm{PPMI} = \max\!\left(0,\ \log_2 \dfrac{O}{E}\right)\)
負值在語料不夠大時不可靠，所以直接忽略；PPMI 常用於詞向量。Negative values are unreliable unless the corpus is huge, so they are dropped. PPMI is common in word vectors.
簡單指標的問題（一）：兩個詞對A problem with simple measures, part 1: two word pairs
\(E(\text{coll.} \mid \text{target}) = P(\text{coll.}) \times \text{count}(\text{target})\)
\(N = 1{,}000{,}000\)。Evert 在這個例子沒有用視窗，而是數雙詞組（bigram）：搭配詞 Y 就是緊鄰在目標詞 X 前面的那個詞。每個 X 前面只有 1 個位置，所以位置數就是 count(X)。N = 1,000,000. Evert does not use a window here but counts bigrams: the collocate Y is the word right before the target X. Each X has exactly one such slot, so the number of slots is just count(X).
the Iliad《伊利亞特》
搭配詞 Y ＝ the：count = 100,000
目標詞 X ＝ Iliad：count = 10
\(P(\text{the}) = \frac{100{,}000}{1{,}000{,}000} = 0.1\)
\(E = 0.1 \times 10 = \mathbf{1}\)
O = 10
目標詞 X ＝ Iliad：count = 10
\(P(\text{the}) = \frac{100{,}000}{1{,}000{,}000} = 0.1\)
\(E = 0.1 \times 10 = \mathbf{1}\)
O = 10
must also「必須」「也」
搭配詞 Y ＝ must：count = 1,000
目標詞 X ＝ also：count = 1,000
\(P(\text{must}) = \frac{1{,}000}{1{,}000{,}000} = 0.001\)
\(E = 0.001 \times 1000 = \mathbf{1}\)
O = 10
目標詞 X ＝ also：count = 1,000
\(P(\text{must}) = \frac{1{,}000}{1{,}000{,}000} = 0.001\)
\(E = 0.001 \times 1000 = \mathbf{1}\)
O = 10
兩個詞對的 O 和 E 完全一樣：O ＝ 10，E ＝ 1。Both pairs have exactly the same O and E.
簡單指標的問題（二）：分數相同A problem with simple measures, part 2: identical scores
| 詞對pair | O | E | O / E | MI |
|---|---|---|---|---|
| the Iliad | 10 | 1 | 10 | 3.32 |
| must also | 10 | 1 | 10 | 3.32 |
分數相同，但情況不同：Iliad 只出現 10 次，而且每一次前面都是 the（O ＝ 10 已是最大可能值）；must、also 各出現 1,000 次，本來可以共現多得多。Same score, different situations: Iliad occurs only 10 times and every one follows “the” (O = 10 is the maximum possible), while must and also occur 1,000 times each and could have co-occurred far more.
O 與 E 忽略了語料中其他的一切——我們還要看兩詞「沒有」共現的情況。O vs. E ignores everything else in the corpus. We also need to look at when the words do not co-occur.
列聯表Contingency table: two events, X and Y
| Y 發生Y occurs | Y 不發生Y does not | |
|---|---|---|
| X 發生X occurs | a | b |
| X 不發生X does not | c | d |
每個個案恰好落入四格之一every case falls into exactly one cell
a ＝ X 和 Y 都發生的個案數cases where both X and Y occur
b ＝ 只有 X 發生（Y 沒有）X occurs, Y does not
c ＝ 只有 Y 發生（X 沒有）Y occurs, X does not
d ＝ 兩者都沒有發生neither X nor Y occurs
X、Y 可以是任何事件；表格也記錄了「都沒有」的情況。詞語只是其中一例（下一頁）。X and Y can be any events, and the table also records when neither happens. Words are just one example (next slide).
列聯表無處不在Contingency tables are everywhere
兩個「是／否」的性質，各自分成兩類，數一數各格有多少個案Two yes/no properties, cross-tabulated: how many cases fall in each cell?
吸煙 × 肺癌Smoking × lung cancer
吸煙者患肺癌的比例是否更高？Are smokers more likely to have cancer?
示意數字 illustrative
慣用手 × 性別Handedness × sex
左撇子的比例是否因性別而異？Does the share of left-handers differ by sex?
教科書例子 textbook example
好 × 天氣（句子）Two words × sentences
兩個詞是否傾向出現在同一個句子？Do the two words tend to share sentences?
本課的小語料 our toy corpus
結構完全相同：問題都是——兩個性質互相獨立嗎？搭配詞的判斷也是如此。Same structure every time: are the two properties independent? Collocations are no different.
n 選 k（一）：放回 vs. 不放回Background “n choose k”, part 1: with vs. without replacement
ABCD
從這 4 個字母中，依序選出 4 個。順序算不同。Pick 4 letters one after another from these 4. The order counts.
放回（同一字母可再選）：每次都有 4 種選擇With replacement (a letter may be chosen again): 4 choices every time
41st
×42nd
×43rd
×44th
＝256\(4^4\)
不放回（每個字母只能用一次）：每選一個，選擇就少一個Without replacement (each letter only once): one fewer choice after every pick
41st
×32nd
×23rd
×14th
＝24\(4!\)
階乘：\(4! = 4 \times 3 \times 2 \times 1 = 24\)，就是把 4 個字母全部排一遍的方法數。Factorial: \(4! = 4 \times 3 \times 2 \times 1 = 24\) is the number of ways to line up all 4 letters.
n 選 k（二）：只選 2 個（不放回）Background “n choose k”, part 2: only pick 2, without replacement
ABCD
現在只選 2 個，不放回；順序仍然算不同（AB ≠ BA）。Now pick only 2, without replacement; the order still counts (AB ≠ BA).
第 1 個有 4 種選擇，第 2 個剩 3 種：4 choices for the 1st pick, 3 left for the 2nd:
41st
×32nd
＝12AB, AC, AD, BA, …
用階乘寫：4! 除以 (4−2)!，去掉多餘的 ×2×1In factorial notation: divide 4! by (4−2)! to cut off the unneeded “×2×1”
\(\dfrac{4!}{(4-2)!}\)
＝\(4 \times 3 \times \cancel{2 \times 1}\)\(\cancel{2 \times 1}\)
＝ 12
一般式：從 n 個中選 k 個（有順序）＝ \(\dfrac{n!}{(n-k)!}\)General form: ordered picks of k out of n
但 AB 與 BA 被算成「不同」；我們只在乎選了哪兩個字母。→ 下一頁But AB and BA count as different; we only care which two letters were chosen. Next slide.
n 選 k（三）：不計順序，除以 k!Background “n choose k”, part 3: order should not matter
12 種有順序的選法中，每一組字母都出現了 2! ＝ 2×1 ＝ 2 次（AB 與 BA 是同一組）：In the 12 ordered picks, every set of letters appears 2! = 2 times (AB and BA are the same set):
ABBA
ACCA
ADDA
BCCB
BDDB
CDDC
所以再除以 2!：12 ÷ 2 ＝ 6 種選法So divide by 2! once more: 12 ÷ 2 = 6 different choices
\(4!\)\((4-2)!\)\(\div\, 2!\)
＝\(4!\)\(2!\,(4-2)!\)
＝\(24\)\(2 \cdot 2\)＝ 6
\(\dbinom{\textcolor{#1f5fd6}{n}}{\textcolor{#e0570f}{k}} = \dfrac{n!}{k!\,(n-k)!}\)
\(= \dfrac{n!}{(n-k)!} \div k!\)
下一頁的奶茶實驗：\(\binom{8}{4} = \dfrac{8!}{4!\,4!} = 70\) 種選法，只有 1 種全對 → \(p = 1/70\)。Tea experiment (next slide): \(\binom{8}{4} = 70\) ways to pick the 4 milk-first cups; only 1 is fully correct, so \(p = 1/70\).
Fisher 與「奶茶實驗」Fisher and the lady tasting tea (1935)
1890–1962
Muriel Bristol 聲稱能分辨奶茶是「先倒牛奶」還是「先倒茶」。Ronald Fisher 設計了實驗來檢驗。A lady claims she can tell whether the milk or the tea was poured first. Fisher designs a test.
M
T
T
M
M
T
M
T
8 杯外觀相同：4 杯先牛奶 (M)、4 杯先茶 (T)，隨機排列；她要指出哪 4 杯是先牛奶。8 identical-looking cups, 4 milk-first and 4 tea-first, in random order. She must pick the 4 milk-first cups.
全部答對 all correct
\(p = \dfrac{1}{\binom{8}{4}} = \dfrac{1}{70} \approx \mathbf{0.014}\)
答對 3 杯 three correct
\(p = \dfrac{17}{70} \approx \mathbf{0.243}\)
全對時，靠猜的機率只有 1/70，這張表「太不可能」；答對 3 杯則可能只是運氣。這就是 Fisher 精確檢定。Guessing all 4 right has probability 1/70, so the table is very unlikely by chance. 3 of 4 could easily be luck. This is Fisher’s exact test.
一張表的機率怎麼算？How likely is one particular table? (hypergeometric distribution)
| 有 Yhas Y | 無 Yno Y | ||
|---|---|---|---|
| 有 Xhas X | a | b | a+b |
| 無 Xno X | c | d | c+d |
| a+c | b+d | N |
\(p =\)
\(\dbinom{\textcolor{#1f5fd6}{a+b}}{\textcolor{#e0570f}{a}}\)\(\dbinom{\textcolor{#1f5fd6}{c+d}}{\textcolor{#e0570f}{c}}\)
\(\dbinom{\textcolor{#7c3aed}{N}}{\textcolor{#e0570f}{a+c}}\)
\(\textcolor{#7c3aed}{N} = a + b + c + d\)：表中所有個案（項目）的總數。\(\textcolor{#7c3aed}{N}\) is the total number of items in the table.
分子 ①：「有 X」這一列有 \(\textcolor{#1f5fd6}{a+b}\) 個項目，其中 \(\textcolor{#e0570f}{a}\) 個有 Y —— 從 \(a+b\) 個中選出 \(a\) 個。Numerator ①: the “has X” row holds \(\textcolor{#1f5fd6}{a+b}\) items; choose the \(\textcolor{#e0570f}{a}\) of them that have Y.
分子 ②：「無 X」那一列有 \(\textcolor{#1f5fd6}{c+d}\) 個項目，其中 \(\textcolor{#e0570f}{c}\) 個有 Y。兩列的選法相乘。Numerator ②: the other row holds \(\textcolor{#1f5fd6}{c+d}\) items, \(\textcolor{#e0570f}{c}\) of which have Y. The two choices are independent, so we multiply.
分母：全部 \(\textcolor{#7c3aed}{N}\) 個項目中，任意挑 \(\textcolor{#e0570f}{a+c}\) 個當作「有 Y」的所有可能。Denominator: all ways to choose which \(\textcolor{#e0570f}{a+c}\) of the \(\textcolor{#7c3aed}{N}\) items have Y, ignoring the rows.
代入 \(a=8,\ b=2,\ c=2,\ d=8\)（\(N=20\)）：Plug in the numbers:
這張表有多「不可能」？How unlikely is this table?
20 位學生：10 位女生、10 位男生；其中 10 位有在讀書，而讀書的人裡有 8 位是女生。性別與讀書有關嗎？20 students: 10 women, 10 men; 10 are studying, and 8 of those are women. Is studying related to sex? (illustrative numbers)
若兩者無關，純屬偶然，預期讀書的女生只有 5 位。If they were unrelated, we would expect only 5 studying women.
觀察到的表 observed table
更極端 more extreme
最極端 most extreme
Fisher 精確檢定：把觀察到的表與所有更極端的表的機率相加Fisher’s exact test sums the observed table and all more extreme ones
→ 顯著 significant
回到 the Iliad ／ must alsoBack to Evert’s example: what does Fisher’s test say?
兩者：O ＝ 10，E ＝ 1，O/E ＝ 10，MI ＝ 3.32 —— 簡單指標分不出來。Both have O = 10, E = 1, O/E = 10, MI = 3.32: the simple measures cannot separate them.
完整的列聯表：每一項是一個雙詞組（共 1,000,000 個）；列 ＝ 前一個詞是不是 the / must，欄 ＝ 後一個詞是不是 Iliad / also。Full contingency tables: every item is a bigram (1,000,000 in total); row = is the first word the / must, column = is the second word Iliad / also.
the Iliad (count(the) = 100,000 · count(Iliad) = 10)
must also (count(must) = count(also) = 1,000)
the Iliad 的 c ＝ 0：每個 Iliad 前面都是 the。must also 的 c ＝ 990：also 大多沒有 must 在前。For the Iliad c = 0: every Iliad follows “the”. For must also c = 990: most alsos have no must before them.
兩者都「顯著」，但 the Iliad 的 p 小 1000 倍 → 用 p 值排序，the Iliad 排在前面。Both are significant, but the Iliad’s p is 1000 times smaller. Ranking by p-value puts the Iliad first, which O/E and MI cannot do.
兩個不同的問題Two different questions we can ask about a word pair
顯著性Statistical significance
「這會不會只是偶然？」“Could this be chance? How sure are we that the attraction is real?”
取決於：關聯的強度 ＋ 證據的多寡（語料越大，越「顯著」）Depends on the strength of the pattern and on how much evidence we have: a bigger corpus gives more significant results.
Fisher plog-likelihood G²t-score
關聯強度（效應量）Strength of association (effect size)
「兩個詞綁得有多緊？」“How tightly are the two words tied together?”
取決於：比例本身；與語料大小無關Depends only on proportions, not on the size of the corpus.
O / EMIDice · logDice
Evert 把兩者都叫作「關聯測量」(association measures)，但它們回答的是不同的問題：顯著不等於強。Evert calls both kinds “association measures”, but they answer different questions. A significant collocation is not necessarily a strong one.
類比：兩枚不公平的硬幣An analogy: two unfair coins
硬幣 ACoin A: few flips
擲 10 次，正面 9 次 · 10 flips, 9 heads
90%
強度 strength90% 正面 heads→ 非常偏 very biased
顯著性 significance\(p \approx 0.011\)→ 勉強 just about
硬幣 BCoin B: many flips
擲 100,000 次，正面 51,000 次 · 100,000 flips, 51,000 heads
51%
強度 strength51% 正面 heads→ 幾乎公平 almost fair
顯著性 significance\(p \approx 10^{-10}\)→ 極其確定 near certain
B 更「顯著」（我們幾乎確定它不公平），但 A 偏得更「強」。p 值衡量的是把握，不是大小。Coin B is more significant: we are almost certain it is unfair. Coin A is more strongly biased. A p-value measures how sure we are, not how big the effect is.
搭配詞也一樣：高頻詞對像硬幣 B，罕見詞對像硬幣 A。Same with collocations: frequent pairs behave like coin B, rare pairs like coin A.
同樣的比例，更大的語料Same proportions, bigger corpus: what changes?
回到「喝 ／ 茶」（視窗 4 + 4）。把語料放大，但每個詞的比例保持不變。Back to (喝, 茶) with a 4 + 4 window. We make the corpus bigger while every proportion stays the same. Values computed from the full contingency table, as in qhchina.
| 強度 strength | 顯著性 significance | ||||||
|---|---|---|---|---|---|---|---|
| N | O | E | O / E | MI | logDice | G² | Fisher p |
| 5,000 | 3 | 0.40 | 7.46 | 2.90 | 8.87 | 7.7 | 0.006 |
| 10,000 | 6 | 0.80 | 7.46 | 2.90 | 8.87 | 15.4 | 0.0001 |
| 20,000 | 12 | 1.61 | 7.46 | 2.90 | 8.87 | 30.7 | \(3 \times 10^{-8}\) |
| 200,000 | 120 | 16.1 | 7.46 | 2.90 | 8.87 | 307 | \(10^{-68}\) |
| 2,000,000 | 1,200 | 161 | 7.46 | 2.90 | 8.87 | 3,070 | \(10^{-669}\) |
強度完全不變：詞對之間的關係沒有改變。Strength does not move: the relationship between the words is the same.
顯著性一直上升：證據越多，把握越大。Significance keeps growing: more evidence, more certainty.
在數百萬詞的語料裡，幾乎所有詞對都「顯著」。只看 p 值，無法知道哪些搭配真正重要。In a corpus of millions of words, almost every pair is “significant”. The p-value alone cannot tell you which collocations matter.
顯著性測量：對數似然比 G²A significance measure: log-likelihood (G²)
觀察值 observed O
期望值 expected E（若兩詞無關 if unrelated）
四格都比較：觀察值偏離期望值多少？Compare all four cells: how far is the observed table from the expected one?
\(G^2 = 2 \sum_{\text{4 cells}} O \cdot \ln \dfrac{O}{E}\)
\(G^2(\text{喝}, \text{茶}) \approx \mathbf{30.7}\)
\(O \cdot \ln(O/E)\)：偏離越大、次數越多，分數越高。Each cell contributes O · ln(O/E): bigger when O is far from E and when the counts are large.
讀法：G² > 3.84 ≈ p < 0.05；G² > 10.83 ≈ p < 0.001。30.7 → 非常顯著。How to read it: G² above 3.84 corresponds to p < 0.05, above 10.83 to p < 0.001. So 30.7 is highly significant.
它是 Fisher 檢定的快速近似（Dunning 1993），回答的是同一個問題：有多確定，而不是有多強。It is a fast approximation to Fisher’s test (Dunning 1993) and answers the same question: how sure, not how strong.
強度測量：Dice 與 logDiceA strength measure: Dice and logDice
兩詞的出現機會中，有多少比例是「在一起」的？Of all the chances the two words had, what share did they spend together?
\(\text{Dice} = \dfrac{2 \cdot O}{\text{「喝」的機會} + \text{「茶」的次數}}\)
分母＝「喝」周圍的位置數 ＋「茶」出現的次數。值介於 0 與 1 之間。Denominator: slots around 喝 plus occurrences of 茶. Always between 0 and 1.
logDice：把 Dice 換成方便閱讀的刻度logDice: Dice on a more readable scale
\(\text{logDice} = 14 + \log_2 \text{Dice}\)
喝 ／ 茶：for (喝, 茶)
\(\text{Dice} = \dfrac{2 \times 12}{800 + 40} \approx 0.029\)
\(\text{logDice} \approx \mathbf{8.87}\)
最高 14（兩詞總是一起出現）；分數越高，關係越強。Maximum 14 (the words always occur together). The higher the score, the stronger the tie.
再看 the Iliad ／ must also：測量會「意見不一」the Iliad vs. must also, once more: the measures disagree
| O | E | MI | logDice | G² | Fisher p | |
|---|---|---|---|---|---|---|
| the Iliad | 10 | 1 | 3.32 | 1.71 | 46.1 | \(1 \times 10^{-10}\) |
| must also | 10 | 1 | 3.32 | 7.36 | 28.2 | \(1 \times 10^{-7}\) |
顯著性選 the Iliad：每一個 Iliad 前面都是 the，證據非常一致。Significance prefers the Iliad: every single Iliad follows “the”, so the evidence is very consistent.
logDice 選 must also：the 出現 100,000 次，只有 10 次在 Iliad 前面；從 the 的角度看，這個關係很弱。logDice prefers must also: “the” occurs 100,000 times and only 10 of those are before Iliad. From the point of view of “the”, the tie is weak.
沒有「唯一正確」的測量；先想清楚你的研究問題要的是哪一種。There is no single correct measure. Decide which question your research needs answered.
實際操作：先篩選，再排序In practice: filter by significance, rank by strength
from qhchina.analytics import find_collocates df = find_collocates( sentences, target_words="王婆", horizon=5, measures=["log_likelihood", "logDice"], correction="fdr_bh", filters={"max_adjusted_p": 0.05, "min_obs_local": 3}, sort_by="log_dice", )
1
篩選：只保留顯著的詞對（p 值或 G²）Filter: keep only pairs that are unlikely to be chance (p-value or G²)
2
校正：同時檢定上千個詞，總會有一些「偶然顯著」→
correctionCorrect: testing thousands of candidates makes some look significant by luck, hence the correction3
排序：用強度（logDice）排列留下的詞對Rank: order the remaining pairs by strength (logDice)
4
回到文本：用 KWIC 閱讀例句；數字只指出方向，解釋要靠文本Read: go back to the passages (KWIC). The numbers point; the texts explain.
三種定義「相鄰」的方法Three ways to define “near”
Evert：先決定什麼是一個共現項目 (item)，再問每個項目是否含 w₁、是否含 w₂Evert: first define what an item is, then ask whether each item contains w₁ and whether it contains w₂
記號（Evert）：w₁ ＝ 目標詞 X，w₂ ＝ 搭配詞 Y；f₁ ＝ count(w₁)，f₂ ＝ count(w₂)，N ＝ 項目總數。O、E 的定義不變。Notation from here on (Evert): w₁ = target X, w₂ = collocate Y; f₁ = count(w₁), f₂ = count(w₂), N = total number of items.
視窗法Window (surface)
Y 在 X 左右 k 個詞之內Y is within k words of X
天氣那麼好我們
計數單位：詞unit: tokens
句子法Sentence (textual)
X 與 Y 在同一個句子裡X and Y are in the same sentence
好天氣…散步
計數單位：句子unit: sentences
語法關係法Grammatical (syntactic)
X 與 Y 有語法關係（如：形容詞→名詞）X and Y stand in a syntactic relation
好→天氣
計數單位：語法搭配對unit: grammatical pairs
視窗法Window approach: Y within k words of X
視窗horizon = 2 ↑ ↓
X = 好Y = 天氣橙色 O₁₁ ＝ O視窗 window（不跨句 stays within a sentence）
項目：語料中每個詞（不含 w₁ 本身）item: every word token except w₁ itself
第 1 列：落在 w₁ 的視窗內row 1: falls inside a span around w₁
第 1 欄：就是搭配詞 w₂column 1: is the collocate w₂
Fisher 精確檢定 p 值Fisher's exact test p-value
Evert 的原例：視窗法 (hat, roll)Evert’s own example: surface cooccurrence, Fig. 58.1 and 58.8
視窗 (L4, R4)：node 詞 hat 左右各 4 個詞，不跨句。Span (L4, R4) around the node word hat, limited by sentence boundaries.
第 2、3 個視窗各含一個 rolled → O ＝ 2The 2nd and 3rd span each contain roll → O = 2
f₁ ＝ 3（hat）、f₂ ＝ 3（roll）、N ＝ 111 個詞f₁ = 3 (hat), f₂ = 3 (roll), N = 111 tokens
項目＝除 hat 以外的每個詞；第 1 列＝落在 hat 視窗內的詞（20 個）Items: every token except hat. Row 1: tokens inside a span around hat (20).
O₁₁ ＝ 2 就是 O；第 1 欄和 ＝ f₂ ＝ 3；第 2 列和 ＝ N − f₁ ＝ 108O₁₁ = 2 is our O; column 1 sums to f₂ = 3; row 2 sums to N − f₁ = 108
與我們的小語料完全同一套定義Same definitions as our toy corpus
Fisher 精確檢定 p 值Fisher's exact test p-value
句子法Sentence approach: X and Y in the same sentence
X = 好Y = 天氣
項目：每一個句子item: every sentence
第 1 列：句子含 w₁row 1: the sentence contains w₁
第 1 欄：句子含 w₂column 1: the sentence contains w₂
橙色格 O₁₁ ＝ O（共現次數）highlighted cell O₁₁ = O, the cooccurrence count
Fisher 精確檢定 p 值Fisher's exact test p-value
Evert 的原例：句子法 (hat, over)Evert’s own example: textual cooccurrence, Fig. 58.2 and 58.7
項目＝句子（共 5 句，N ＝ 5）Items are sentences; N = 5
hat 在 3 句（f₁ ＝ 3），over 在 2 句（f₂ ＝ 2），同句出現 1 次（O ＝ 1）hat is in 3 sentences, over in 2, both in 1 sentence
最後一句雖有兩個 over，仍只算 1 次The last sentence has two overs but counts once
O₁₁ ＝ 1，O₁₂ ＝ 2，O₂₁ ＝ 1，O₂₂ ＝ 1one sentence has both, two only hat, one only over, one neither
列和 ＝ f₁ ＝ 3，欄和 ＝ f₂ ＝ 2，總和 ＝ N ＝ 5row sum = f₁ = 3, column sum = f₂ = 2, total = N = 5
Fisher 精確檢定 p 值Fisher's exact test p-value
語法關係法Grammatical approach: X and Y in a syntactic relation
只計「形容詞 → 名詞」的修飾關係；謂語用法（心情不好）不算。Only adjective → noun modifiers count; predicative uses are ignored.
找到的搭配對 pairs found
項目：每一個「形容詞→名詞」搭配對item: every adjective → noun pair token
第 1 列：第一個詞是 w₁row 1: the first word is w₁
第 1 欄：第二個詞是 w₂column 1: the second word is w₂
橙色格 O₁₁ ＝ Ohighlighted cell = cooccurrence count
Fisher 精確檢定 p 值Fisher's exact test p-value
Evert 的原例：語法關係法 (young, gentleman)Evert’s own example: syntactic cooccurrence, Fig. 58.3 and 58.6
(O, f₁, f₂, N)
＝ (1, 3, 3, 9)
＝ (1, 3, 3, 9)
項目＝每一個「形容詞→名詞」搭配對（pair token），共 9 個（N ＝ 9）Items: the 9 adjective → noun pair tokens listed on the right
其他所有詞（in, stood, corduroy …）都不算All other words are disregarded
young 修飾 3 個名詞（f₁ ＝ 3），其中 1 個是 gentleman（O ＝ 1）young modifies 3 nouns; one of them is gentleman
gentleman 被 3 個形容詞修飾（f₂ ＝ 3）：stout, old, younggentleman is modified by 3 adjectives: stout, old, young
O₂₂ ＝ 4：與 young、gentleman 都無關的 4 對4 pairs contain neither word
Fisher 精確檢定 p 值Fisher's exact test p-value
同一個問題，三種數法Same question, three ways of counting
| 視窗法Window | 句子法Sentence | 語法關係法Grammatical | |
|---|---|---|---|
| 「相鄰」是指“near” means | k 個詞之內within k words | 同一個句子same sentence | 有語法關係syntactic relation |
| 計數單位unit counted | 詞tokens | 句子sentences | 搭配對pairs |
| 需要needs | 選定 ka choice of k | 分句sentence splitting | 句法分析器a parser |
| (好, 天氣) 的表table for (好, 天氣); orange = O₁₁ = O |
回顧Recap
1
搭配詞：兩個詞比偶然更常在附近一起出現A collocation: two words that occur near each other more often than chance
2
觀察值 O 對 期望值 E；取對數得 MI，去掉負值得 PPMIObserved O vs. expected E; the log gives MI, dropping negatives gives PPMI
3
列聯表：也記錄「沒有」共現的情況（a, b, c, d）Contingency table: also records the cases where the words do not co-occur
4
Fisher 精確檢定：這張表在「純屬偶然」下有多不可能？Fisher’s exact test: how unlikely is this table if it were all chance?
5
顯著性 ≠ 強度：p 值、G² 問「有多確定」；MI、logDice 問「有多強」Significance is not strength: p and G² ask how sure we are, MI and logDice ask how strong the tie is
6
三種「相鄰」：視窗、句子、語法關係 —— 同一張表，不同的「項目」Three kinds of “near”: window, sentence, grammatical: same table, different items
7
下一步：在真實語料上計算，比較不同的指標Next: compute these on a real corpus and compare the measures