"Гока" meaning in Русский

See Гока in All languages combined, or Wiktionary

Noun

Etymology: Происходит от ??
  1. женское имя
    Sense id: ru-Гока-ru-noun-ArmtoQtp
The following are not (yet) sense-disambiguated
Hypernyms: имя
{
  "anagrams": [
    {
      "word": "Агко"
    },
    {
      "word": "Кога"
    }
  ],
  "categories": [
    {
      "kind": "other",
      "name": "Женские имена/ru",
      "parents": [],
      "source": "w"
    },
    {
      "kind": "other",
      "name": "Женский род/ru",
      "parents": [],
      "source": "w"
    },
    {
      "kind": "other",
      "name": "Имена собственные",
      "parents": [],
      "source": "w"
    },
    {
      "kind": "other",
      "name": "Нет сведений о составе слова/ru",
      "parents": [],
      "source": "w"
    },
    {
      "kind": "other",
      "name": "Нужна классификация по Зализняку/ru",
      "parents": [],
      "source": "w"
    },
    {
      "kind": "other",
      "name": "Нужна этимология",
      "parents": [],
      "source": "w"
    },
    {
      "kind": "other",
      "name": "Одушевлённые/ru",
      "parents": [],
      "source": "w"
    },
    {
      "kind": "other",
      "name": "Русские лексемы",
      "parents": [],
      "source": "w"
    },
    {
      "kind": "other",
      "name": "Русские существительные",
      "parents": [],
      "source": "w"
    },
    {
      "kind": "other",
      "name": "Русский язык",
      "parents": [],
      "source": "w"
    },
    {
      "kind": "other",
      "name": "Слова из 4 букв/ru",
      "parents": [],
      "source": "w"
    }
  ],
  "etymology_text": "Происходит от ??",
  "hypernyms": [
    {
      "sense_index": 1,
      "word": "имя"
    }
  ],
  "hyphenations": [
    {
      "parts": [
        "Гока"
      ]
    }
  ],
  "lang": "Русский",
  "lang_code": "ru",
  "pos": "noun",
  "senses": [
    {
      "examples": [
        {
          "bold_text_offsets": [
            [
              18,
              22
            ]
          ],
          "text": "Мою подругу зовут Гока."
        }
      ],
      "glosses": [
        "женское имя"
      ],
      "id": "ru-Гока-ru-noun-ArmtoQtp"
    }
  ],
  "tags": [
    "animate"
  ],
  "word": "Гока"
}
{
  "anagrams": [
    {
      "word": "Агко"
    },
    {
      "word": "Кога"
    }
  ],
  "categories": [
    "Женские имена/ru",
    "Женский род/ru",
    "Имена собственные",
    "Нет сведений о составе слова/ru",
    "Нужна классификация по Зализняку/ru",
    "Нужна этимология",
    "Одушевлённые/ru",
    "Русские лексемы",
    "Русские существительные",
    "Русский язык",
    "Слова из 4 букв/ru"
  ],
  "etymology_text": "Происходит от ??",
  "hypernyms": [
    {
      "sense_index": 1,
      "word": "имя"
    }
  ],
  "hyphenations": [
    {
      "parts": [
        "Гока"
      ]
    }
  ],
  "lang": "Русский",
  "lang_code": "ru",
  "pos": "noun",
  "senses": [
    {
      "examples": [
        {
          "bold_text_offsets": [
            [
              18,
              22
            ]
          ],
          "text": "Мою подругу зовут Гока."
        }
      ],
      "glosses": [
        "женское имя"
      ]
    }
  ],
  "tags": [
    "animate"
  ],
  "word": "Гока"
}

Download raw JSONL data for Гока meaning in Русский (0.9kB)


This page is a part of the kaikki.org machine-readable Русский dictionary. This dictionary is based on structured data extracted on 2025-08-01 from the ruwiktionary dump dated 2025-07-21 using wiktextract (ed078bd and 3c020d2). 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.