'How can I remove the thousand comma separator when converting data frame columns? [duplicate]
Given the following data frame:
State,City,Population,Poverty_Rate,Median_Age,
VA,XYZ,.,10.5%,42,
MD,ABC,"12,345",8.9%,.,
NY,.,987,654,.,41,
...
import pandas as pd
df = pd.read_csv("/path... /sample_data")
df.dtypes returns
State Object
City Object
Population Object
Proverty_Rate Object
Median_Age Object
I attempt to convert the data type of appropriate columns to int or float:
df = df.astype({"Population": int, "Proverty_rate": float, "Median_Age": int })
I received
Value Error: invalid literal for int() with base 10: '12,345'
I suspect the comma separator is causing this problem. How can I remove those from my dataset?
Solution 1:[1]
There is an argument in Pandas DataFrame as pd.read_csv(thousands=',') which is set to None by default.
data = """
State City Population Poverty_Rate Median_Age
VA XYZ 500,00 10.5% 42
MD ABC 12,345 8.9% .
NY . 987,654 . 41"""
from io import StringIO
import pandas as pd
df = pd.read_csv(StringIO(data),sep='\s+',thousands=',')
print(df)
State City Population Poverty_Rate Median_Age
0 VA XYZ 50000 10.5% 42
1 MD ABC 12345 8.9% .
2 NY . 987654 . 41
Ideally, what you need to do is replace the string markers and then coerce your string columns into integers/floats.
#using your dict.
int_cols = ({"Population": int, "Poverty_Rate": float, "Median_Age": int })
for col in int_cols.keys():
df[col] = pd.to_numeric(df[col].astype(str).str.replace('%',''),errors='coerce')
print(df.dtypes)
State object
City object
Population int64
Poverty_Rate float64
Median_Age float64
dtype: object
print(df)
State City Population Poverty_Rate Median_Age
0 VA XYZ 50000 10.5 42.0
1 MD ABC 12345 8.9 NaN
2 NY . 987654 NaN 41.0
Solution 2:[2]
Could you try the following? First do a str.replace on the column before you cast it to an integer?
import pandas as pd
df = pd.DataFrame([
{'value': '123,445'},
{'value': '143,445,788'}
])
df['value'] = df['value'].str.replace(',', '').astype(int)
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | Dilini Peiris |
| Solution 2 | JQadrad |
