2 回答

TA貢獻(xiàn)1909條經(jīng)驗(yàn) 獲得超7個(gè)贊
堆棧和 與 on ,默認(rèn)情況下刪除 :groupby+agg.joinlevel=0stack()NaN
df1['patient address'] = df.stack().groupby(level=0).agg(' '.join)
0 a 202
1 b x 201
2 c 203
dtype: object

TA貢獻(xiàn)1860條經(jīng)驗(yàn) 獲得超8個(gè)贊
import numpy as np
import pandas as pd
df = pd.DataFrame(
{
"Addr1": ["a", "b", "c"],
"Addr2": ["", "x", np.nan],
"Addr3": ["202", "201", "203"],
}
)
df1 = pd.DataFrame(
{
"patient address": df[["Addr1", "Addr2", "Addr3"]]
.fillna(" ")
.agg(" ".join, axis=1)
}
)
print(df1)
輸出:
patient address
0 a 202
1 b x 201
2 c 203
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