"poeta" meaning in 拉丁語

See poeta in All languages combined, or Wiktionary

Noun

Audio: la-cls-poeta.ogg Forms: poētae
Etymology: 来自古希臘語 ποιητής (poiētḗs, “詩人,作家”)。
  1. 詩人
    Sense id: zh-poeta-la-noun-iEcy~OwZ
The following are not (yet) sense-disambiguated
Derived forms: poētaster, poētor Related terms: poēma Coordinate_terms: poētria
{
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      "name": "拉丁語名詞",
      "parents": [],
      "source": "w"
    },
    {
      "kind": "other",
      "name": "拉丁語第一類變格名詞",
      "parents": [],
      "source": "w"
    },
    {
      "kind": "other",
      "name": "拉丁語第一類變格陽性名詞",
      "parents": [],
      "source": "w"
    },
    {
      "kind": "other",
      "name": "拉丁語詞元",
      "parents": [],
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    },
    {
      "kind": "other",
      "name": "拉丁語陽性名詞",
      "parents": [],
      "source": "w"
    },
    {
      "kind": "other",
      "name": "有3個詞條的頁面",
      "parents": [],
      "source": "w"
    },
    {
      "kind": "other",
      "name": "派生自古希臘語的拉丁語詞",
      "parents": [],
      "source": "w"
    },
    {
      "kind": "other",
      "name": "源自古希臘語的拉丁語借詞",
      "parents": [],
      "source": "w"
    }
  ],
  "coordinate_terms": [
    {
      "word": "poētria"
    }
  ],
  "derived": [
    {
      "word": "poētaster"
    },
    {
      "word": "poētor"
    }
  ],
  "etymology_text": "来自古希臘語 ποιητής (poiētḗs, “詩人,作家”)。",
  "forms": [
    {
      "form": "poētae",
      "raw_tags": [
        "属格"
      ]
    }
  ],
  "lang": "拉丁語",
  "lang_code": "la",
  "pos": "noun",
  "related": [
    {
      "word": "poēma"
    }
  ],
  "senses": [
    {
      "glosses": [
        "詩人"
      ],
      "id": "zh-poeta-la-noun-iEcy~OwZ"
    }
  ],
  "sounds": [
    {
      "audio": "la-cls-poeta.ogg",
      "mp3_url": "https://upload.wikimedia.org/wikipedia/commons/transcoded/3/38/La-cls-poeta.ogg/La-cls-poeta.ogg.mp3",
      "ogg_url": "https://commons.wikimedia.org/wiki/Special:FilePath/la-cls-poeta.ogg",
      "raw_tags": [
        "音频(古典)"
      ]
    }
  ],
  "tags": [
    "masculine"
  ],
  "word": "poeta"
}
{
  "categories": [
    "拉丁語名詞",
    "拉丁語第一類變格名詞",
    "拉丁語第一類變格陽性名詞",
    "拉丁語詞元",
    "拉丁語陽性名詞",
    "有3個詞條的頁面",
    "派生自古希臘語的拉丁語詞",
    "源自古希臘語的拉丁語借詞"
  ],
  "coordinate_terms": [
    {
      "word": "poētria"
    }
  ],
  "derived": [
    {
      "word": "poētaster"
    },
    {
      "word": "poētor"
    }
  ],
  "etymology_text": "来自古希臘語 ποιητής (poiētḗs, “詩人,作家”)。",
  "forms": [
    {
      "form": "poētae",
      "raw_tags": [
        "属格"
      ]
    }
  ],
  "lang": "拉丁語",
  "lang_code": "la",
  "pos": "noun",
  "related": [
    {
      "word": "poēma"
    }
  ],
  "senses": [
    {
      "glosses": [
        "詩人"
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  ],
  "sounds": [
    {
      "audio": "la-cls-poeta.ogg",
      "mp3_url": "https://upload.wikimedia.org/wikipedia/commons/transcoded/3/38/La-cls-poeta.ogg/La-cls-poeta.ogg.mp3",
      "ogg_url": "https://commons.wikimedia.org/wiki/Special:FilePath/la-cls-poeta.ogg",
      "raw_tags": [
        "音频(古典)"
      ]
    }
  ],
  "tags": [
    "masculine"
  ],
  "word": "poeta"
}

Download raw JSONL data for poeta meaning in 拉丁語 (0.9kB)

{
  "called_from": "extractor/zh/page/parse_section/192",
  "msg": "Unhandled subtitle: 派生语汇",
  "path": [
    "poeta"
  ],
  "section": "拉丁語",
  "subsection": "派生语汇",
  "title": "poeta",
  "trace": ""
}

This page is a part of the kaikki.org machine-readable 拉丁語 dictionary. This dictionary is based on structured data extracted on 2024-11-03 from the zhwiktionary dump dated 2024-10-20 using wiktextract (d6bf104 and a5af179). The data shown on this site has been post-processed and various details (e.g., extra categories) removed, some information disambiguated, and additional data merged from other sources. See the raw data download page for the unprocessed wiktextract data.

If you use this data in academic research, please cite Tatu Ylonen: Wiktextract: Wiktionary as Machine-Readable Structured Data, Proceedings of the 13th Conference on Language Resources and Evaluation (LREC), pp. 1317-1325, Marseille, 20-25 June 2022. Linking to the relevant page(s) under https://kaikki.org would also be greatly appreciated.