在`str.format()`中使用多个if条件
在`str.format()`中使用多个if条件
提问于 2018-12-17 02:52:59
我有一个类似下面的数据框架。
我检查了得分1、2、3列，并打印了各自的主题。我可以比较两列并打印相应的文本。
如何包含不同的列？
import pandas as pd
import numpy as np
raw_data = {'Sub1':['A','B','C','D','E'],
'Sub2':['F','G','H','I','J'],
'Sub3':['K','L','M','N','O'],
'S_score1': [1, 0, 0, 6,0],
'F_score1' : [0, 1,0,0,0],
'L_score1' : [1,2,3,0,4],
'S_score2': [0, 0, 0, 6,0],
'F_score2' : [0, 1,0,0,0],
'L_score2' : [1,2,3,0,4],
'S_score3': [0, 0, 0, 6,0],
'F_score3' : [0, 1,0,0,0],
'L_score3' : [1,2,3,0,4]}
df2 = pd.DataFrame(raw_data, columns = ['Sub1','Sub2','Sub3','S_score1', 'F_score1','L_score1','S_score2', 'F_score2','L_score2','S_score3', 'F_score3','L_score3'])
def S_text(f):
s_text = "You have scored on {}" .format(f['Sub1']) if f['S_score1'] >= 1 else "You have scored on {}" .format(f['Sub2'])
return s_text
def F_text(f):
f_text = "You have scored on {}" .format(f['Sub1']) if f['F_score1'] >= 1 else "You have scored on {}" .format(f['Sub2'])
return f_text
def L_text(f):
l_text = "You have scored on {}" .format(f['Sub1']) if f['L_score1'] >= 1 else "You have scored on {}" .format(f['Sub2'])
return l_text
df2['s_text'] = df2.apply(S_text, axis=1)
df2['f_text'] = df2.apply(F_text, axis=1)
df2['l_text'] = df2.apply(L_text, axis=1)我看起来像下面的类型比较，但这给出了错误。基本上，我希望如果两列满足我想要打印的两个主题的条件(scores>=1)。如果3列满足条件(scores>=1)，并且我想打印文本中的3个主题，如下所示，2个条件。有没有其他方法可以比较3列并打印文本。
def S_text(f):
s_text = "You have scored on {}" .format(f['Sub1']) if f['S_score1'] >= 1
elif f['S_score2'] >= 1 "You have scored on {}" .format(f['Sub2'])
elif f['S_score3'] >=1 "You have scored on {}" .format(f['Sub3'])
elif f['S_score1'] >=1 and f['S_score2']>=1 "You have scored on {} {}" .format(f['Sub1'], f['Sub2'])
elif f['S_score1'] >=1 and f['S_score3']>=1 "You have scored on {} {}" .format(f['Sub1'], f['Sub3'])
elif f['S_score2'] >=1 and f['S_score3']>=1 "You have scored on {} {}" .format(f['Sub2'], f['Sub3'])
elif f['S_score1'] >=1 and f['S_score2']>=1 and f['S_score3']>=1 "You have scored on {} {} {}" .format(f['Sub1'],f['Sub2'], f['Sub3'])
return s_text想要的输出：
回答 1
Stack Overflow用户
发布于 2018-12-17 07:52:22
您的df与预期结果不匹配，它应该是这样的，以匹配您的输出：
raw_data = {'Sub1':['A','B','C','D','E'],
'Sub2':['F','G','H','I','J'],
'Sub3':['K','L','M','N','O'],
'S_score1': [1, 0, 0, 6,0],
'F_score1' : [0, 1,0,0,0],
'L_score1' : [1,2,3,0,4],
'S_score2': [0, 1, 0, 6,0],
'F_score2' : [0, 1,0,0,0],
'L_score2' : [1,2,3,0,4],
'S_score3': [0, 1, 0, 6,0],
'F_score3' : [0, 1,0,0,0],
'L_score3' : [1,2,3,0,4]}
df2 = pd.DataFrame(raw_data, columns = ['Sub1','Sub2','Sub3','S_score1', 'F_score1','L_score1','S_score2', 'F_score2','L_score2','S_score3', 'F_score3','L_score3'])我添加了一个循环来处理不同的标签。这个时间与输出相匹配。
import re
letters = ['S', 'F', 'L']
for letter in letters:
for row in range(0, len(df2)):
df_ = df2[[letter+'_score1', letter+'_score2', letter+'_score3']].iloc[row:row+1,0:3]
row_string = ''
for col in df_.columns:
if df_[col][row] >= 1:
row_string = row_string + ', '+ str(df2.iloc[row][df_.columns.get_loc(col)])
row_string = re.sub(r'^,', '', row_string)
if row_string == '':
df2.loc[row:,letter+'_text'] = 'You have not Scored any subject'
else:
df2.loc[row:,letter+'_text'] = 'You have scored on' + row_string
display(df2))
Sub1 Sub2 Sub3 S_score1 F_score1 L_score1 S_score2 F_score2 L_score2 S_score3 F_score3 L_score3 S_text F_text L_text
0 A F K 1 0 1 0 0 1 0 0 1 You have scored on A You have not Scored any subject You have scored on A, F, K
1 B G L 0 1 2 1 1 2 1 1 2 You have scored on G, L You have scored on B, G, L You have scored on B, G, L
2 C H M 0 0 3 0 0 3 0 0 3 You have not Scored any subject You have not Scored any subject You have scored on C, H, M
3 D I N 6 0 0 6 0 0 6 0 0 You have scored on D, I, N You have not Scored any subject You have not Scored any subject
4 E J O 0 0 4 0 0 4 0 0 4 You have not Scored any subject You have not Scored any subject You have scored on E, J, O我相信也有更简单的方法可以做到这一点。
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原文链接：
https://stackoverflow.com/questions/53805431
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