About Leizhou
Leizhou is a large city located in People's Republic of China, in the Guangdong region. According to the latest available data (2026), the population of Leizhou is approximately 1,321,091 residents, making it the #351 most populous city in People's Republic of China.
Globally, Leizhou ranks #1,317 among the world's most populated cities. With over 1 million residents, Leizhou plays an important role as an economic and cultural center in People's Republic of China.
Leizhou City Statistics
| City | Leizhou |
| Country | |
| Population | 1,321,091 |
| Global Rank | #1,317 |
| Rank in People's Republic of China | #351 of 2,667 cities |
| City Category | large city |
| Region | Guangdong |
| Latitude | 20.9193 |
| Longitude | 110.0789 |
| Data Year | 2026 |
Population Comparison
To put Leizhou's population of 1,321,091 into perspective:
- Leizhou has approximately 1.3 million residents.
- This makes it comparable in size to cities like Prague or Munich.
Where is Leizhou
Located?
Leizhou is located in People's Republic of China, in the Guangdong region, at coordinates 20.9193° latitude and 110.0789° longitude. Located near the equator, Leizhou experiences a tropical climate with warm temperatures throughout the year.
Nearby Cities
Other Cities in People's Republic of China
Largest Cities in People's Republic of China
| # | City | Population |
|---|---|---|
| 1 | Kaifeng | 4,824,016 |
| 2 | Yuncheng | 4,774,508 |
| 3 | Chenzhou | 4,667,134 |
| 4 | Yibin | 4,588,804 |
| 5 | Huai'an | 4,556,230 |
Frequently Asked Questions
What is the population of Leizhou?
The population of Leizhou is approximately 1,321,091 residents according to the latest available data (2026).
Where is Leizhou located?
Leizhou is located in People's Republic of China at approximately 20.9193° latitude and 110.0789° longitude.
Is Leizhou one of the largest cities in People's Republic of China?
With a population of 1,321,091, Leizhou is one of the major urban areas in People's Republic of China. It ranks #351 out of 2,667 cities in the country.