I am a beginner in pandas and python, asking for a little bit help. Here is my dataset, the k_symbol column labels either UVER or SIPO, I want to replace UVER as int 0, and SIPO as int 1.
I tried dff.replace(to_replace=['k_symbol'], value=[1, 0]), but it does not seem right. Appreciate for any suggestions
apply() functions are notoriously slow, so if you care about speed, consider one of these solutions
df["k_symbol"].map({"UVER":0, "SIPO":1})
df["k_symbol"] = (df["k_symbol"] == "SIPO").astype(int)
%%timeit
df["k_symbol"] = (df["k_symbol"] == "SIPO").astype(int)
10 loops, best of 3: 83.3 ms per loop
%%timeit
df['k_symbol'].apply(lambda x : 0 if x == 'UVER' else 1 )
1 loop, best of 3: 550 ms per loop
%%timeit
df["k_symbol"].map({"UVER":0,"SIPO":1})
10 loops, best of 3: 83.6 ms per loop
Use this single line to get desired result.
df.k_symbol = df.k_symbol.apply(lambda o : 1 if o == 'SIPO' else 0 if o == 'UVER' else o)
You can simplify it as below , if all other than SIPO will be 0
df.k_symbol = df.k_symbol.apply(lambda o : 1 if o == 'SIPO' else 0)
import pandas as pd
df = pd.DataFrame(["SIPO","UVER"] * 3, columns=["k_symbol"])
df["k_symbol"].map({"UVER":0,"SIPO":1})
Output: df
k_symbol
0 SIPO
1 UVER
2 SIPO
3 UVER
4 SIPO
5 UVER
mapped:
0 1
1 0
2 1
3 0
4 1
5 0
Use .loc :
import pandas as pd
df = pd.DataFrame(
[[1, "SIPO"], [0, "UVER"], [0, "UVER"], [0, "UVER"], [1, "UVER"],],
columns=["gender", "k_symbol"],
)
df.loc[df["k_symbol"] == "SIPO", "k_symbol"] = 1
df.loc[df["k_symbol"] == "UVER", "k_symbol"] = 0
print(df)
Returning:
gender k_symbol
0 1 1
1 0 0
2 0 0
3 0 0
4 1 0
You can pass an anonymous function (Lambda) stating a condition to check within apply.
df['k_symbol'] = df['k_symbol'].apply(lambda x : 0 if x == 'UVER' else 1 )
I believe a better (faster) way is using .eq():
df['k_symbol'] = df['k_symbol'].eq('SIPO').astype(int)
