我有两个df,一个是我的几个ID的列表,另一个是人名和ID。
我想让他们循环,当df1中的ID等于df2 ID时,他在df2中取名字并在df1中创建。
我试图以我发现但没有创建的混乱状态来适应此代码。
for key,row in df.iterrows():
choices = str(list(df2.NAME_ID.unique()))
names = process.extract(str(row['P1_ID']), choices, limit=2)[0][0]
name = df2[df2['NAME_ID'] == names]['NAME']
if not name.empty:
df.loc[key,'Name'] = name
import pandas as pd
df = pd.read_clipboard(sep='\s\s+')
GAME_DATE_EST GAME_ID GAME_STATUS_TEXT P1_ID P2_ID SEASON P1_ID PTS_P1
0 2020-01-01 21900504 Final 1610612764 1610612753 2019 1610612764 10
1 2020-01-01 21900505 Final 1610612752 1610612757 2019 1610612752 9
2 2020-01-01 21900506 Final 1610612749 1610612750 2019 1610612749 10
3 2020-01-01 21900507 Final 1610612747 1610612756 2019 1610612747 8
4 2019-12-31 21900497 Final 1610612766 1610612738 2019 1610612766 9
df2
NAME_ID STANDINGSDATE NAME G W L W_PCT
0 1610612747 2020-01-01 Math 34 27 7 0.79
1 1610612743 2020-01-01 John 33 23 10 0.70
2 1610612746 2020-01-01 Elias 35 24 11 0.69
3 1610612745 2020-01-01 Alexander 34 23 11 0.68
4 1610612742 2020-01-01 Michael 33 21 12 0.64
希望您能理解并能帮助我
答案 0 :(得分:1)
为此,您可以执行一个简单的join
:
.grid-container {
align-items: start;
display: grid;
grid-gap: 10px;
}
.grid-column-count-2 {
grid-template-columns: repeat(2, minmax(0, 1fr));
}
.grid-row-count-2 {
grid-template-rows: repeat(2, min-content);
}
.col-1 {
grid-column-start: 1;
}
.col-2 {
grid-column-start: 2;
}
.row-1 {
grid-row-start: 1;
}
.row-2 {
grid-row-start: 2;
}
.rowspan-2 {
grid-row-end: span 2;
}
.align-self-end {
align-self: end;
}
.align-self-stretch {
align-self: stretch;
}
根据您提供的数据,您可以尝试:
<div class="grid-container grid-column-count-2 grid-row-count-2">
<div class="col-1 row-1 rowspan-2 align-self-stretch" style="background-color: #f2f2f2; border: 1px solid #bbb; overflow-y: scroll;"><span>div A</span><br>
<span>div B</span><br><span>div C</span><br><span>div D</span><br></div>
<input type="text" class="col-2 row-1" />
<input type="text" class="col-2 row-2 align-self-end" />
<input type="text" class="col-2 row-1" />
</div>