遍历相同元素的列表

Question

我有以下函数，它返回pandas系列的States - Associated Counties

def answer():
    census_df.set_index(['STNAME', 'CTYNAME'])
    for name, state, cname in zip(census_df['STNAME'], census_df['STATE'], census_df['CTYNAME']):
        print(name, state, cname)

Alabama 1 Tallapoosa County
Alabama 1 Tuscaloosa County
Alabama 1 Walker County
Alabama 1 Washington County
Alabama 1 Wilcox County
Alabama 1 Winston County
Alaska 2 Alaska
Alaska 2 Aleutians East Borough
Alaska 2 Aleutians West Census Area
Alaska 2 Anchorage Municipality
Alaska 2 Bethel Census Area
Alaska 2 Bristol Bay Borough
Alaska 2 Denali Borough
Alaska 2 Dillingham Census Area
Alaska 2 Fairbanks North Star Borough

我想知道其中县最多的州。我可以像这样遍历每个状态：

counter = 0
counter2 = 0
for name, state, cname in zip(census_df['STNAME'], census_df['STATE'], census_df['CTYNAME']):
     if state == 1:
         counter += 1
 print(counter)
     if state == 1:
         counter2 += 1
 print(counter2)

等等。我可以设定状态数量 (rng = range(1, 56)) 并遍历它，但创建 56 个列表是一场噩梦。如果这样做有没有更简单的方法？

Answer 1

Pandas 允许我们在没有循环/迭代的情况下执行此类操作：

In [21]: df.STNAME.value_counts()
Out[21]:
Alaska     9
Alabama    6
Name: STNAME, dtype: int64

In [24]: df.STNAME.value_counts().head(1)
Out[24]:
Alaska    9
Name: STNAME, dtype: int64

或

In [18]: df.groupby('STNAME')['CTYNAME'].count()
Out[18]:
STNAME
Alabama    6
Alaska     9
Name: CTYNAME, dtype: int64

In [19]: df.groupby('STNAME')['CTYNAME'].count().idxmax()
Out[19]: 'Alaska'