See Подопрыгорова on 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 masculin singulier de Подопрыгоров." ], "id": "fr-Подопрыгорова-ru-name-WnbrBt6T" }, { "glosses": [ "Génitif masculin singulier de Подопрыгоров." ], "id": "fr-Подопрыгорова-ru-name-WYUYfAmP" }, { "glosses": [ "Nominatif féminin singulier de Подопрыгоров." ], "id": "fr-Подопрыгорова-ru-name-FJRVs4yp" } ], "tags": [ "form-of", "surename" ], "word": "Подопрыгорова" }
{ "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 masculin singulier de Подопрыгоров." ] }, { "glosses": [ "Génitif masculin singulier de Подопрыгоров." ] }, { "glosses": [ "Nominatif féminin singulier de Подопрыгоров." ] } ], "tags": [ "form-of", "surename" ], "word": "Подопрыгорова" }
Download raw JSONL data for Подопрыгорова meaning in All languages combined (0.5kB)
This page is a part of the kaikki.org machine-readable All languages combined dictionary. This dictionary is based on structured data extracted on 2025-01-24 from the frwiktionary dump dated 2025-01-20 using wiktextract (c15a5ce and 5c11237). 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.