# Unemployment: San Lorenzo Municipio, PR (U)

Unemployment: San Lorenzo Municipio, PR (U) was down 4.2% year on year in September 2026, compared with 13.8% in August 2026.
The level stood at 1,291.0 thousands.
That is the lowest reading since December 2024.
Over the past ten years it has ranged from −36.2% (Nov 2018) to 79.6% (Dec 2025), averaging 2.7%.

- Source agency: U.S. Bureau of Labor Statistics (flat files)
- Makrognosis code: `US-PR-72129.UNEMPLVL.M.NSA`
- Frequency: monthly
- Units: thousands
- Coverage: 1990-01-01 to 2026-09-01 (439 observations)
- Page: https://makrognosis.robomacro.com/data/united-states/unemployment-san-lorenzo-municipio-pr-u

## Recent data

| Period | Value | Y/Y % |
|---|---:|---:|
| September 2026 | 1,291.0 thousands | −4.2 |
| August 2026 | 1,686.0 thousands | 13.8 |
| July 2026 | 1,736.0 thousands | 31.7 |
| June 2026 | 1,620.0 thousands | 20.8 |
| May 2026 | 1,368.0 thousands | 22.5 |
| April 2026 | 1,309.0 thousands | 17.5 |
| March 2026 | 1,342.0 thousands | 28.7 |
| February 2026 | 1,357.0 thousands | 28.6 |
| January 2026 | 1,452.0 thousands | 42.1 |
| December 2025 | 1,403.0 thousands | 79.6 |
| November 2025 | 1,347.0 thousands | 70.7 |
| October 2025 | 1,354.0 thousands | 67.6 |
| September 2025 | 1,348.0 thousands | 69.8 |
| August 2025 | 1,482.0 thousands | 74.6 |
| July 2025 | 1,318.0 thousands | 43.9 |
| June 2025 | 1,341.0 thousands | 58.7 |
| May 2025 | 1,117.0 thousands | 42.3 |
| April 2025 | 1,114.0 thousands | 45.2 |
| March 2025 | 1,043.0 thousands | 29.1 |
| February 2025 | 1,055.0 thousands | 22.0 |
| January 2025 | 1,022.0 thousands | 13.2 |
| December 2024 | 781.0 thousands | −6.7 |
| November 2024 | 789.0 thousands | −2.1 |
| October 2024 | 808.0 thousands | −0.1 |

## API

```
curl -H "X-API-Key: $MAKROGNOSIS_KEY" \
  "https://makrognosis.robomacro.com/v1/series/US-PR-72129.UNEMPLVL.M.NSA"

# Python
import requests
r = requests.get("https://makrognosis.robomacro.com/v1/series/US-PR-72129.UNEMPLVL.M.NSA",
                 headers={"X-API-Key": KEY})
obs = r.json()["observations"]
```