See Загребаевых in All languages combined, or Wiktionary
{
"categories": [
{
"kind": "other",
"name": "Formes de noms de famille en russe",
"parents": [],
"source": "w"
},
{
"kind": "other",
"name": "Russe",
"orig": "russe",
"parents": [],
"source": "w"
}
],
"lang": "Russe",
"lang_code": "ru",
"pos": "name",
"pos_title": "Forme de nom de famille",
"senses": [
{
"glosses": [
"Accusatif pluriel de Загребаев."
],
"id": "fr-Загребаевых-ru-name-1Q0~kB2i"
},
{
"glosses": [
"Génitif pluriel de Загребаев."
],
"id": "fr-Загребаевых-ru-name-CrHlBscN"
},
{
"glosses": [
"Locatif pluriel de Загребаев."
],
"id": "fr-Загребаевых-ru-name-NTuF-H9o"
}
],
"tags": [
"form-of",
"surename"
],
"word": "Загребаевых"
}
{
"categories": [
"Formes de noms de famille en russe",
"russe"
],
"lang": "Russe",
"lang_code": "ru",
"pos": "name",
"pos_title": "Forme de nom de famille",
"senses": [
{
"glosses": [
"Accusatif pluriel de Загребаев."
]
},
{
"glosses": [
"Génitif pluriel de Загребаев."
]
},
{
"glosses": [
"Locatif pluriel de Загребаев."
]
}
],
"tags": [
"form-of",
"surename"
],
"word": "Загребаевых"
}
Download raw JSONL data for Загребаевых meaning in Russe (0.4kB)
This page is a part of the kaikki.org machine-readable Russe dictionary. This dictionary is based on structured data extracted on 2025-07-19 from the frwiktionary dump dated 2025-07-01 using wiktextract (45c4a21 and f1c2b61). 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.