当我运行以下代码时，出现以下错误：ValueError：int（）以10为底的无效文字：“（（1，0，'Friday'）”

Question

When I run the following code,I get this

ValueError: invalid literal for int()  with base 10: "(1, 0, 'Friday')" 

pointing to the line:

monthwise = list(map(int, read_from_csv(infile_csv, month=True)[0])) .

I have included the sample .csv file

Output: Plot I need that compares monthly readership among subscribers and customers

import calendar  
import datetime 
infile_csv = 'C:/pythonscripts/NYC-2016-Summary.csv'   
def read_from_csv(input_csvfile, duration=False, month=False, hour=False, day_of_week=False):    

# assign columns name
if duration==True:
    col_name='duration'
elif month==True:
    col_name='month'
elif hour==True:
    col_name='hour'
elif day_of_week==True:
    col_name='day_of_week'

# init lists for output
n_ridership4column = []
n_ridership_sub = []
n_ridership_cust = []

with open(infile_csv, 'r') as f_in:
    filereader = csv.DictReader(f_in)
    for row in filereader:
        n_ridership4column.append(row[col_name])
        if row['user_type'] == 'Subscriber':
            n_ridership_sub.append(row[col_name])
        else:
            n_ridership_cust.append(row[col_name])

return n_ridership4column, n_ridership_sub, n_ridership_cust

# using the function above to get monthly ridership

monthwise = list(map(int, read_from_csv(infile_csv, month=True)[0]))   
monthwise_sub = list(map(int, read_from_csv(infile_csv, month=True)[1])) 
monthwise_cust = list(map(int, read_from_csv(infile_csv, month=True)[2]))  

The below code is for plotting. This is not required for the question but for the clarity of output.

fig, ax = plt.subplots()
bins = [i for i in range(1,14)]                 
#upper bound is 14 to accomodate bin for december

#### Plotting monthly total along with customers and subscribers stacked

ax.hist(monthwise, bins=bins, edgecolor='k', align='left', label='Total Ridership', stacked= True)   
ax.hist(monthwise_sub, bins=bins, edgecolor='k', align='left', label='Subscribers', stacked=True)
ax.hist(monthwise_cust, bins=bins, edgecolor='k', align='left', label='Customer', stacked=True)

ax.set_xticks(bins[:-1])
ax.set_xticklabels(list(calendar.month_abbr[i] for i in bins[:-1]))


plt.title('Monthly Ridership in NYC', fontsize=16)
plt.xlabel('Monthly', fontsize=14)  
plt.ylabel('Rides', fontsize=14)
plt.xticks(fontsize=12)
plt.yticks(fontsize=12)

plt.legend()
plt.show()

This is my sample .csv file for the above code

duration    month   hour    day_of_week user_type
13.98333333 (1, 0, 'Friday')        (1, 0, 'Friday')    Customer
11.43333333 (1, 0, 'Friday')        (1, 0, 'Friday')    Subscriber
5.25    (1, 0, 'Friday')        (1, 0, 'Friday')    Subscriber
12.31666667 (1, 0, 'Friday')        (1, 0, 'Friday')    Subscriber
20.88333333 (1, 0, 'Friday')        (1, 0, 'Friday')    Customer
8.75    (1, 0, 'Friday')        (1, 0, 'Friday')    Subscriber
10.98333333 (1, 0, 'Friday')        (1, 0, 'Friday')    Subscriber
7.733333333 (1, 1, 'Friday')        (1, 1, 'Friday')    Subscriber
3.433333333 (1, 1, 'Friday')        (1, 1, 'Friday')    Subscriber
7.083333333 (1, 1, 'Friday')        (1, 1, 'Friday')    Customer
13.3    (1, 2, 'Friday')        (1, 2, 'Friday')    Subscriber
9.733333333 (1, 2, 'Friday')        (1, 2, 'Friday')    Subscriber
8.416666667 (1, 2, 'Friday')        (1, 2, 'Friday')    Subscriber

Answer 1

The error message means you are trying to parse a value which isn't numeric as an integer. When you ask Python to do something it can't do (divide a number by zero, reference an undeclared variable, etc) it throws an error. Usually the error message is clear enough, though when you are only just learning Python, you sometimes need to google.

To the extent that blame can be allocated, whichever program wrote this broken pseudo-CSV is wrong and should be fixed or replaced. For CSV to be useful, it needs to be nomalized to one datum per field, though you see this principle violated from time to time. Writing a composite field in a Python-specific format is at the very least misdirected, and in this case quite probably a bug.

In addition, there is one column less in the sample data than the sample headers suggest. On the other hand, columns 2 and 3 always seem to be identical, and vaguely appear to be composed of values which would fit the apparently expected values for columns 2, 3, and 4 in the headers.

Your code is weird in that it seems to read the file every time it wants to extract a column. This might vaguely make sense if your input file is too huge to fit into memory all at once; but in the absence of any such concerns in your question or in comments within the code, I would recommend reading all the columns into memory once. This should also make your program at least an order of magnitude faster.

The DictReader already takes care of collecting its input into an OrderedDict so the append in the loop is simply duplicating work which this Python library is already performing for you.

Perhaps something like this would suit your needs if you are stuck with this broken CSV.

def parse_broken_csv(filename):
    rows = []
    with open(filename, 'r') as fh:
        reader = csv.reader(fh, delimiter='\t')
        headers = reader.__next__()
        for duration, tpl, _, cust in reader:
            month, hour, dow = tpl.strip('()').split(', ')
            rows.append({
                'duration': float(duration),
                'month': int(month),
                'hour': int(hour),
                'day_of_week': dow.strip("'"),
                'cust': cust})
    return rows

rows = parse_broken_csv(r'NYC-2016-Summary.csv')

monthwise = [row['month'] for row in rows]
monthwise_sub = [row['month'] for row in rows if row['cust'] == 'Subscriber']
monthwise_cust = [row['month'] for row in rows if row['cust'] == 'Customer']

For the sample CSV you posted, the value of rows is

[
 {'duration': 13.98333333, 'month': 1, 'day_of_week': 'Friday', 'cust': 'Customer', 'hour': 0},
 {'duration': 11.43333333, 'month': 1, 'day_of_week': 'Friday', 'cust': 'Subscriber', 'hour': 0},
 {'duration': 5.25, 'month': 1, 'day_of_week': 'Friday', 'cust': 'Subscriber', 'hour': 0},
 {'duration': 12.31666667, 'month': 1, 'day_of_week': 'Friday', 'cust': 'Subscriber', 'hour': 0},
 {'duration': 20.88333333, 'month': 1, 'day_of_week': 'Friday', 'cust': 'Customer', 'hour': 0},
 {'duration': 8.75, 'month': 1, 'day_of_week': 'Friday', 'cust': 'Subscriber', 'hour': 0},
 {'duration': 10.98333333, 'month': 1, 'day_of_week': 'Friday', 'cust': 'Subscriber', 'hour': 0}
]

and the value of monthwise is

[1, 1, 1, 1, 1, 1, 1]