在Pandas数据帧中查找第一个匹配项

是的,你可以使用 Series.item 如果查找始终从 Series :

westcoast.loc[westcoast.state=='Oregon', 'capital'].item()

s = westcoast.loc[westcoast.state=='Oregon', 'capital']
s = np.nan if s.empty else s.iat[0] 
print (s) #Salem

s = westcoast.loc[westcoast.state=='New York', 'capital']
s = np.nan if s.empty else s.iat[0] 
print (s)
nan

处理异常的更通用解决方案,因为有3种可能的输出场景:

westcoast = pd.DataFrame([['Washington','Olympia'],['Oregon','Salem'],
                          ['California','Sacramento'],['Oregon','Portland']],
                        columns=['state','capital'])

print (westcoast)
        state     capital
0  Washington     Olympia
1      Oregon       Salem
2  California  Sacramento
3      Oregon    Portland

s = westcoast.loc[westcoast.state=='Oregon', 'capital']

#if not value returned
if s.empty:
    s = 'no match'

#if only one value returned
elif len(s) == 1:
    s = s.item()
else:

# if multiple values returned, return a list of values
    s = s.tolist()

print (s)
['Salem', 'Portland']

可以创建查找函数:

def look_up(a):
    s = westcoast.loc[westcoast.state==a, 'capital']
    #for no match
    if s.empty:
        return np.nan
    #for match only one value
    elif len(s) == 1:
        return s.item()
    else:
    #for return multiple values
        return s.tolist()

print (look_up('Oregon'))
['Salem', 'Portland']

print (look_up('California'))
Sacramento

print (look_up('New Yourk'))
nan

state 索引:

state_capitals = westcoast.set_index('state')['capital']
print(state_capitals['Oregon'])
# Salem

通过索引,每个查找都是 O(1) 平均而言 westcoast['state']=='Oregon' O(n) 比较。当然,建立索引也是 O(n) ,因此您需要进行多次查找才能获得回报。

同时,一旦你 state_capitals 州首府 .