Зейнсвилл in All languages combined

IPA: ˈzʲeɪ̯nsvʲɪɫ [Show additional information ▼] [Hide additional information ▲]

Etymology: От ?? Forms: Зе́йнсвилл [nominative, singular], Зе́йнсвиллы [nominative, plural], Зе́йнсвилла [genitive, singular], Зе́йнсвиллов [genitive, plural], Зе́йнсвиллу [dative, singular], Зе́йнсвиллам [dative, plural], Зе́йнсвилл [accusative, singular], Зе́йнсвиллы [accusative, plural], Зе́йнсвиллом [instrumental, singular], Зе́йнсвиллами [instrumental, plural], Зе́йнсвилле [prepositional, singular], Зе́йнсвиллах [prepositional, plural]

[Show JSON for postprocessed kaikki.org data shown on this page ▼] [Hide JSON for postprocessed kaikki.org data shown on this page ▲]

{
  "categories": [
    {
      "kind": "other",
      "name": "Мужской род/ru",
      "parents": [],
      "source": "w"
    },
    {
      "kind": "other",
      "name": "Неодушевлённые/ru",
      "parents": [],
      "source": "w"
    },
    {
      "kind": "other",
      "name": "Нет сведений о составе слова/ru",
      "parents": [],
      "source": "w"
    },
    {
      "kind": "other",
      "name": "Нужна этимология",
      "parents": [],
      "source": "w"
    },
    {
      "kind": "other",
      "name": "Русские лексемы",
      "parents": [],
      "source": "w"
    },
    {
      "kind": "other",
      "name": "Русские существительные",
      "parents": [],
      "source": "w"
    },
    {
      "kind": "other",
      "name": "Русские существительные, склонение 1a",
      "parents": [],
      "source": "w"
    },
    {
      "kind": "other",
      "name": "Русский язык",
      "parents": [],
      "source": "w"
    },
    {
      "kind": "other",
      "name": "Слова из 9 букв/ru",
      "parents": [],
      "source": "w"
    },
    {
      "kind": "other",
      "name": "Топонимы/en",
      "parents": [],
      "source": "w"
    },
    {
      "kind": "other",
      "name": "Требуется категоризация/ru",
      "parents": [],
      "source": "w"
    }
  ],
  "etymology_text": "От ??",
  "forms": [
    {
      "form": "Зе́йнсвилл",
      "tags": [
        "nominative",
        "singular"
      ]
    },
    {
      "form": "Зе́йнсвиллы",
      "tags": [
        "nominative",
        "plural"
      ]
    },
    {
      "form": "Зе́йнсвилла",
      "tags": [
        "genitive",
        "singular"
      ]
    },
    {
      "form": "Зе́йнсвиллов",
      "tags": [
        "genitive",
        "plural"
      ]
    },
    {
      "form": "Зе́йнсвиллу",
      "tags": [
        "dative",
        "singular"
      ]
    },
    {
      "form": "Зе́йнсвиллам",
      "tags": [
        "dative",
        "plural"
      ]
    },
    {
      "form": "Зе́йнсвилл",
      "tags": [
        "accusative",
        "singular"
      ]
    },
    {
      "form": "Зе́йнсвиллы",
      "tags": [
        "accusative",
        "plural"
      ]
    },
    {
      "form": "Зе́йнсвиллом",
      "tags": [
        "instrumental",
        "singular"
      ]
    },
    {
      "form": "Зе́йнсвиллами",
      "tags": [
        "instrumental",
        "plural"
      ]
    },
    {
      "form": "Зе́йнсвилле",
      "tags": [
        "prepositional",
        "singular"
      ]
    },
    {
      "form": "Зе́йнсвиллах",
      "tags": [
        "prepositional",
        "plural"
      ]
    }
  ],
  "lang": "Русский",
  "lang_code": "ru",
  "pos": "noun",
  "senses": [
    {
      "glosses": [
        "город в США"
      ],
      "id": "ru-Зейнсвилл-ru-noun-jF3p~O9b"
    }
  ],
  "sounds": [
    {
      "ipa": "ˈzʲeɪ̯nsvʲɪɫ"
    }
  ],
  "tags": [
    "declension-2",
    "inanimate",
    "masculine"
  ],
  "translations": [
    {
      "lang": "Английский",
      "lang_code": "en",
      "word": "Zanesville"
    }
  ],
  "word": "Зейнсвилл"
}

[Show JSON for raw wiktextract data ▼] [Hide JSON for raw wiktextract data ▲]

{
  "categories": [
    "Мужской род/ru",
    "Неодушевлённые/ru",
    "Нет сведений о составе слова/ru",
    "Нужна этимология",
    "Русские лексемы",
    "Русские существительные",
    "Русские существительные, склонение 1a",
    "Русский язык",
    "Слова из 9 букв/ru",
    "Топонимы/en",
    "Требуется категоризация/ru"
  ],
  "etymology_text": "От ??",
  "forms": [
    {
      "form": "Зе́йнсвилл",
      "tags": [
        "nominative",
        "singular"
      ]
    },
    {
      "form": "Зе́йнсвиллы",
      "tags": [
        "nominative",
        "plural"
      ]
    },
    {
      "form": "Зе́йнсвилла",
      "tags": [
        "genitive",
        "singular"
      ]
    },
    {
      "form": "Зе́йнсвиллов",
      "tags": [
        "genitive",
        "plural"
      ]
    },
    {
      "form": "Зе́йнсвиллу",
      "tags": [
        "dative",
        "singular"
      ]
    },
    {
      "form": "Зе́йнсвиллам",
      "tags": [
        "dative",
        "plural"
      ]
    },
    {
      "form": "Зе́йнсвилл",
      "tags": [
        "accusative",
        "singular"
      ]
    },
    {
      "form": "Зе́йнсвиллы",
      "tags": [
        "accusative",
        "plural"
      ]
    },
    {
      "form": "Зе́йнсвиллом",
      "tags": [
        "instrumental",
        "singular"
      ]
    },
    {
      "form": "Зе́йнсвиллами",
      "tags": [
        "instrumental",
        "plural"
      ]
    },
    {
      "form": "Зе́йнсвилле",
      "tags": [
        "prepositional",
        "singular"
      ]
    },
    {
      "form": "Зе́йнсвиллах",
      "tags": [
        "prepositional",
        "plural"
      ]
    }
  ],
  "lang": "Русский",
  "lang_code": "ru",
  "pos": "noun",
  "senses": [
    {
      "glosses": [
        "город в США"
      ]
    }
  ],
  "sounds": [
    {
      "ipa": "ˈzʲeɪ̯nsvʲɪɫ"
    }
  ],
  "tags": [
    "declension-2",
    "inanimate",
    "masculine"
  ],
  "translations": [
    {
      "lang": "Английский",
      "lang_code": "en",
      "word": "Zanesville"
    }
  ],
  "word": "Зейнсвилл"
}

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This page is a part of the kaikki.org machine-readable All languages combined dictionary. This dictionary is based on structured data extracted on 2024-10-06 from the ruwiktionary dump dated 2024-10-02 using wiktextract (9f93753 and c1a3a36). 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.

"Зейнсвилл" meaning in All languages combined

Noun [Русский]