无法在python3中更改布尔变量的值

时间:2019-12-09 09:11:58

标签: python python-3.x pandas

我不知道为什么我的代码中'add_command'变量的值始终为'True'。 请帮我。这是我的代码:

df3 = xls.parse(1)

# Function finding will return 4 arrays
Spec3, Grade3, other_scrap3, months3 = finding(df3)

# if Spec3[4] == df1.loc[1, 'Spec'] and Grade3[4] == df1.loc[1, 'Material']:
#     print('Equal')

add_command = False                  # Variable to confirm need to add row for df1 or not
# Looping all row of data frame df1 then find out whether Spec3[i] & Grade3[i] in df1 or not.
for i in range(0, len(Spec3)):
    date, mon = split_month(months3[i])
    for ind, row in df1.iterrows():
        if row['Spec'] == Spec3[i] and row['Material'] == Grade3[i]:
           df1.loc[ind, mon] = other_scrap3[i]
           add_command = False
        else:
           add_command = True
    print(add_command)
    if add_command == True:
       df2 = pd.DataFrame(columns = column_names)
       df2.loc[0, 'Spec'] = Spec3[i]
       df2.loc[0, 'Material'] = Grade3[i]
       df2.loc[0, mon] = other_scrap3[i]
       df1 = df1.append(df2, ignore_index=True)           
       print(df1)

请注意:在第3行和第4行中,我可以打印“等于”,这意味着add_commande应该从True更改为False。 但实际上在第14行:“ print(add_command)”,我看到此变量始终为'True'。 非常感谢您的帮助!

**EDIT 1:**  
Here is an example of df1:    
        Spec Material  Jan  Feb  Mar  Apr  May  Jun  Jul  Aug  Sep  Oct  Nov  
0        NaN      NaN  NaN  NaN  NaN  NaN  NaN  NaN  NaN  NaN  NaN  NaN  NaN     
1  40x25x2.0   SPHT-4  NaN  NaN  NaN  NaN  NaN  NaN  NaN  NaN   52  NaN  NaN     
2   42.7x3.5   SPHT-1  NaN  NaN  NaN  NaN  NaN  NaN  NaN  NaN    8  NaN  NaN     
3   42.7x1.6   SPHT-3  NaN  NaN  NaN  NaN  NaN  NaN  NaN  NaN    8  NaN  NaN     
4   42.7x2.0   SPHT-4  NaN  NaN  NaN  NaN  NaN  NaN  NaN  NaN    4  NaN  NaN     
5   22.2x3.2   SPHT-1  NaN  NaN  NaN  NaN  NaN  NaN  NaN  NaN   20  NaN  NaN     
6   48.6x3.5   SPHT-1  NaN  NaN  NaN  NaN  NaN  NaN  NaN  NaN    8  NaN  NaN     
7   42.0x1.0  HFS436L  NaN  NaN  NaN  NaN  NaN  NaN  NaN  NaN   14  NaN  NaN    

这是Spec3和Grade3数组:

Spec3:  
['60x47.6x2.0', '60x47.6x2.0', '60.5x2.6', '60.5x2.0', '40x25x2.0', '25.4x3.2']  
Grade3:  
['SPHT2-D15M', 'SPHT-4', 'SPHT-3', 'SPHT-1', 'SPHT-4', 'SPHT2-D15M']

编辑2: 我只删除了else: add_command = True并将add_command = False移到for i in range(0, len(Spec)):内,然后就可以了。 无法以逻辑方式解释其工作原理。 这是修改后的代码:

df3 = xls.parse(1)

# Function finding will return 4 arrays
Spec3, Grade3, other_scrap3, months3 = finding(df3)

# if Spec3[4] == df1.loc[1, 'Spec'] and Grade3[4] == df1.loc[1, 'Material']:
#     print('Equal')

#add_command = False                  **# Delete this line**
# Looping all row of data frame df1 then find out whether Spec3[i] & Grade3[i] in df1 or not.
for i in range(0, len(Spec3)):
    add_command = True                  # Add this line
    date, mon = split_month(months3[i])
    for ind, row in df1.iterrows():
        if row['Spec'] == Spec3[i] and row['Material'] == Grade3[i]:
           df1.loc[ind, mon] = other_scrap3[i]
           add_command = False
        #else:                             **# Delete this line**
           #add_command = True             **# Delete this line**
    print(add_command)
    if add_command == True:
       df2 = pd.DataFrame(columns = column_names)
       df2.loc[0, 'Spec'] = Spec3[i]
       df2.loc[0, 'Material'] = Grade3[i]
       df2.loc[0, mon] = other_scrap3[i]
       df1 = df1.append(df2, ignore_index=True)           
       print(df1)

2 个答案:

答案 0 :(得分:0)

if条件从不成立,因此从未达到add_command = False。您可以打印变量以查看运行时值,例如IF语句之前的print('{}\t{}\t{}\t{}'.format(row['Spec'], Spec3[i], row['Material'],Grade3[i])

如果您知道一组应该为您提供IF真实条件的值,则还可以通过数据框操作对其进行检查。

答案 1 :(得分:0)

好的,这是你的答案

for i in range(0, len(Spec3)): #you loop for each index in spec3
    add_command = True                  # Add this line
    date, mon = split_month(months3[i])
    for ind, row in df1.iterrows(): #then you loop through the rows 
        if row['Spec'] == Spec3[i] and row['Material'] == Grade3[i]:
           df1.loc[ind, mon] = other_scrap3[i]
           add_command = False
           # Here it changes to False once it found the condition in the row that is why it is working 
        #else:                             **# Delete this line**
           #add_command = True             **if you have the add_command =True**
           # For each item the add_command will be different it will change all the time
           # so the add command will give the value for the last item all the time

    print(add_command)
    if add_command == True:
       df2 = pd.DataFrame(columns = column_names)
       df2.loc[0, 'Spec'] = Spec3[i]
       df2.loc[0, 'Material'] = Grade3[i]
       df2.loc[0, mon] = other_scrap3[i]
       df1 = df1.append(df2, ignore_index=True)           
       print(df1)