# Labor Force: Davenport-Moline-Rock Island, IA-IL Metropolitan Statistical Area (U)

Labor Force: Davenport-Moline-Rock Island, IA-IL Metropolitan Statistical Area (U) was down 0.2% year on year in September 2025, compared with −0.4% in August 2025.
The level stood at 188,252 thousands.
Over the past ten years it has ranged from −6.8% (Jul 2020) to 2.5% (Jan 2025), averaging −0.1%.

- Source agency: U.S. Bureau of Labor Statistics (flat files)
- Makrognosis code: `US-19340.LF.M.NSA`
- Frequency: monthly
- Units: thousands
- Coverage: 1990-01-01 to 2026-09-01 (440 observations)
- Page: https://makrognosis.robomacro.com/data/united-states/labor-force-davenport-moline-rock-island-ia-il-metropolitan-statistical

## Recent data

| Period | Value | Y/Y % |
|---|---:|---:|
| September 2026 | 188,252 thousands | — |
| August 2026 | 187,413 thousands | — |
| July 2026 | 190,017 thousands | — |
| June 2026 | 190,672 thousands | — |
| May 2026 | 188,066 thousands | — |
| April 2026 | 188,296 thousands | — |
| March 2026 | 190,918 thousands | — |
| February 2026 | 190,579 thousands | — |
| January 2026 | 191,179 thousands | — |
| December 2025 | 189,758 thousands | — |
| November 2025 | 191,749 thousands | — |
| September 2025 | 192,559 thousands | −0.2 |
| August 2025 | 191,392 thousands | −0.4 |
| July 2025 | 192,053 thousands | 0.05 |
| June 2025 | 193,753 thousands | 1.5 |
| May 2025 | 191,256 thousands | 1.1 |
| April 2025 | 191,660 thousands | 1.9 |
| March 2025 | 191,482 thousands | 1.4 |
| February 2025 | 191,306 thousands | 1.5 |
| January 2025 | 191,924 thousands | 2.5 |
| December 2024 | 190,936 thousands | 1.0 |
| November 2024 | 192,172 thousands | 0.8 |
| October 2024 | 192,148 thousands | 0.5 |
| September 2024 | 192,934 thousands | 0.7 |

## API

```
curl -H "X-API-Key: $MAKROGNOSIS_KEY" \
  "https://makrognosis.robomacro.com/v1/series/US-19340.LF.M.NSA"

# Python
import requests
r = requests.get("https://makrognosis.robomacro.com/v1/series/US-19340.LF.M.NSA",
                 headers={"X-API-Key": KEY})
obs = r.json()["observations"]
```