Python 基于另一列计算一列中的重复值

我使用Panda处理以下CSV数据类型：

f,f,f,f,f,t,f,f,f,t,f,t,g,f,n,f,f,t,f,f,f,f,f,f,f,f,f,f,f,f,f,f,f,t,t,t,nowin
t,f,f,f,f,f,f,f,f,f,t,f,g,f,b,f,f,t,f,f,f,f,f,t,f,t,f,f,f,f,f,f,f,t,f,n,won
t,f,f,f,t,f,f,f,t,f,t,f,g,f,b,f,f,t,f,f,f,t,f,t,f,t,f,f,f,f,f,f,f,t,f,n,won
f,f,f,f,f,f,f,f,f,f,t,f,g,f,b,f,f,t,f,f,f,f,f,t,f,t,f,f,f,f,f,f,f,t,f,n,nowin
t,f,f,f,t,f,f,f,t,f,t,f,g,f,b,f,f,t,f,f,f,t,f,t,f,t,f,f,f,f,f,f,f,t,f,n,won
f,f,f,f,f,f,f,f,f,f,t,f,g,f,b,f,f,t,f,f,f,f,f,t,f,t,f,f,f,f,f,f,f,t,f,n,win

对于这部分原始数据，我试图返回如下内容：

Column1_name -- t -- counts of nowin = 0

Column1_name -- t -- count of wins = 3

Column1_name -- f -- count of nowin = 2 

Column1_name -- f -- count of win = 1

基于这个想法，我想做这样的事情：

print(df[df.target == 'won'].count())

然而，这将基于最后一列返回相同数量的“WON”，而不考虑此列是“f”还是“t”。在其他例子中，我希望使用Panda dataframe工作中的一些东西，从SQL中产生“groupby”的概念，例如，基于第一列和最后一列进行分组

我应该继续坚持这样的想法吗？我应该简单地开始使用for循环吗

如果需要，请参阅我的其余代码：

import pandas as pd


url = "https://archive.ics.uci.edu/ml/machine-learning-databases/chess/king-rook-vs-king-pawn/kr-vs-kp.data"

df = pd.read_csv(url,names=[
                       'bkblk','bknwy','bkon8','bkona','bkspr','bkxbq','bkxcr','bkxwp','blxwp','bxqsq','cntxt','dsopp','dwipd',
                        'hdchk','katri','mulch','qxmsq','r2ar8','reskd','reskr','rimmx','rkxwp','rxmsq','simpl','skach','skewr',
                        'skrxp','spcop','stlmt','thrsk','wkcti','wkna8','wknck','wkovl','wkpos','wtoeg','target'
                        ])


features = ['bkblk','bknwy','bkon8','bkona','bkspr','bkxbq','bkxcr','bkxwp','blxwp','bxqsq','cntxt','dsopp','dwipd',
        'hdchk','katri','mulch','qxmsq','r2ar8','reskd','reskr','rimmx','rkxwp','rxmsq','simpl','skach','skewr',
        'skrxp','spcop','stlmt','thrsk','wkcti','wkna8','wknck','wkovl','wkpos','wtoeg','target']


# number of lines 
#tot_of_records = np.size(my_data,0) 
#tot_of_records = np.unique(my_data[:,1])

#for item in my_data:
#    item[:,0]
num_of_won=0
num_of_nowin=0

for item in df.target:
    if item == 'won':
        num_of_won = num_of_won + 1
    else:
        num_of_nowin = num_of_nowin + 1

print(num_of_won)
print(num_of_nowin)        

print(df[df.target == 'won'].count())  

#print(df[:1])
#print(df.bkblk.to_string(index=False))
#print(df.target.unique())
#ini_entropy = (() + ())

这可能会起作用-

outdf=df.apply（lambda x:pd.crosstab（index=df.target，columns=x）。to_dict（））

基本上，我们将进入每个功能列，并与目标列建立一个交叉表

希望这有帮助！：）