如何以表格形式打印此词典？

Question

i have a defaultdict like this:

{('Montag', '17.30'): [True, False], ('Dienstag', '16.30'): [True, False], ('Mittwoch', '15.30'): [True, False], ('Donnerstag', '14.30'): [True, False, False, True], ('Freitag', '13.30'): [True, False], ('Samstag', '12.30'): [True, False], ('Sonntag', '11.30'): [True, False], ('Sonntag', '17.30'): [False, True], ('Samstag', '16.30'): [False, True], ('Freitag', '15.30'): [False, True], ('Mittwoch', '13.30'): [False, True], ('Dienstag', '12.30'): [False, True], ('Montag', '11.30'): [False, True], ('Donnerstag', '16.30'): [False, True], ('Samstag', '11.25'): [True,True]})

and i want to print this in form of a table like this:

Montag Dienstag Mittwoch Donnerstag Freitag Samstag Sonntag  
0      0        0        0          0       100     0        11.25
50     0        0        0          0       0       50       11.30
0      50       0        0          0       50      0        12.30
0      0        50       0          50      0       0        13.30
0      0        0        50         0       0       0        14.30
0      0        50       0          50      0       0        15.30
0      50       0        50         0       50      0        16.30
50     0        0        0          0       0       50       17.30

On the x-axis I want to output all days that occur in the dict next to each other. (each day only once)

On the Y-axis every time which occurs in the dict should be output to each other.

The table should be filled with the ratio of False and True (maybe with statistics.mean()).

I only solved to print the axis with this code:

WOCHENTAGE = {0: "Montag",
             1: "Dienstag",
              2: "Mittwoch",
              3: "Donnerstag",
              4: "Freitag",
              5: "Samstag",
              6: "Sonntag"}

set_for_day = set()
set_for_time = set()
for k, v in testdict.items():
    set_for_day.add(k[0])
    set_for_time.add(k[1])

order = list(WOCHENTAGE.values())    
for day in sorted(set_for_day, key = lambda x: order.index(x)):
    print(f"{day} ", end ="")
print()
for times in sorted(set_for_time):
    print(f"                                                            {times}")

Answer 1

The main challenge here is the format in which the data is given. The (day,time) tuple as the key to the dict makes it difficult to index the dict to get the wanted value for each day/time combination. As shown in the code below, this can be fixed by transforming the data into a dict which can be indexed as data[day][time] , returning the percentage of true values. Using a defaultdict , which you already mentioned in your question, avoids having to fill in zeros for missing values.

Computing the percentage given a list of boolean values can be done using sum : each True is counted as one, and each False as zero. Divide by the length to get the mean, and multiply by 100 to get the percentage. I used sum(bool(v) for v in lst) in case some non-bool values (like integers) are passed in. If you want, you can change it to just sum(lst) .

The output of the code below matches your desired output.

from collections import defaultdict

# The example data.
data = {
    ('Montag', '17.30'): [True, False],
    ('Dienstag', '16.30'): [True, False],
    ('Mittwoch', '15.30'): [True, False],
    ('Donnerstag', '14.30'): [True, False, False, True],
    ('Freitag', '13.30'): [True, False],
    ('Samstag', '12.30'): [True, False],
    ('Sonntag', '11.30'): [True, False],
    ('Sonntag', '17.30'): [False, True],
    ('Samstag', '16.30'): [False, True],
    ('Freitag', '15.30'): [False, True],
    ('Mittwoch', '13.30'): [False, True],
    ('Dienstag', '12.30'): [False, True],
    ('Montag', '11.30'): [False, True],
    ('Donnerstag', '16.30'): [False, True],
    ('Samstag', '11.25'): [True,True]
}

# Week days, in order.
WEEK_DAYS = [
    "Montag",
    "Dienstag",
    "Mittwoch",
    "Donnerstag",
    "Freitag",
    "Samstag",
    "Sonntag"
]

# Given a list of values, return the percentage that are truthy.
def percentage_true(lst):
    return 100 * sum(bool(v) for v in lst) / len(lst)


# The list of days and times present in the data.
present_days = list(set(k[0] for k in data.keys()))
present_times = list(set(k[1] for k in data.keys()))

# Sort these days based on WEEK_DAYS.
present_days.sort(key = WEEK_DAYS.index)
# Sort the times by converting to minutes.
present_times.sort(key = lambda s: 60 * int(s[:2]) + int(s[3:]))

# Re-organize the data such that it can be indexed as
# data[day][time] => percentage. Use a defaultdict to
# return 0 for absent values.
data = {
    day: defaultdict(lambda: 0, {
        k[1]: percentage_true(v)
        for k, v in data.items() if k[0] == day
    })
    for day in set(k[0] for k in data.keys())
}

# Print the header.
for day in present_days:
    print(day, end=" ")
print()

# For printing, find the lengths of the day names, and the
# formats required for .format().
day_lengths = [len(s) for s in present_days]
perc_formats = ["{{:<{}.0f}}".format(l) for l in day_lengths]

# Print the values row-by-row.
for time in present_times:
    for day, fmt in zip(present_days, perc_formats):
        print(fmt.format(data[day][time]), end=" ")
    print(time)

Answer 2

Try this:

data = {('Montag', '17.30'): [True, False], ('Dienstag', '16.30'): [True, False], ('Mittwoch', '15.30'): [True, False], ('Donnerstag', '14.30'): [True, False, False, True], ('Freitag', '13.30'): [True, False], ('Samstag', '12.30'): [True, False], ('Sonntag', '11.30'): [True, False], ('Sonntag', '17.30'): [False, True], ('Samstag', '16.30'): [False, True], ('Freitag', '15.30'): [False, True], ('Mittwoch', '13.30'): [False, True], ('Dienstag', '12.30'): [False, True], ('Montag', '11.30'): [False, True], ('Donnerstag', '16.30'): [False, True], ('Samstag', '11.25'): [True,True]}


# get unique times
time = []
for item in data:
    time.append(item[1])

time_list = list(set(time))
sorted_time = sorted(time_list, key=float)

# set days
days = ['Montag', 'Dienstag', 'Mittwoch', 'Donnerstag', 'Freitag', 'Samstag', 'Sonntag']

# create data sets
proper_data = []
for time in sorted_time:
    for day in days:
        for key, value in data.items():
            if key[1] == time and key[0] == day:
                if value.count(True) == 1:
                    proper_data.append('50')
                elif value.count(True) == 2:
                    proper_data.append('100')
        else:
            proper_data.append('0')
    proper_data.append(time)


# remove additional items
item_indexes = [n+1 for n,x in enumerate(proper_data) if x=='50' or x =='100']

for index in sorted(item_indexes, reverse=True):
    del proper_data[index]

# slice data into parts
final_data = []
for i in range(int(len(proper_data)/8)):
    final_data.append(proper_data[(8*i):(8*(i+1))])


# add time to names
days.append('Time')

# print data
final_data = [days] + final_data
for item in final_data:
    print("{:<10}{:<10}{:<10}{:<10}{:<10}{:<10}{:<10}{:<10}".format(item[0], item[1], item[2], item[3], item[4], item[5], item[6], item[7]))

Output:

Montag    Dienstag  Mittwoch  DonnerstagFreitag   Samstag   Sonntag   Time      
0         0         0         0         0         100       0         11.25     
50        0         0         0         0         0         50        11.30     
0         50        0         0         0         50        0         12.30     
0         0         50        0         50        0         0         13.30     
0         0         0         100       0         0         0         14.30     
0         0         50        0         50        0         0         15.30     
0         50        0         50        0         50        0         16.30     
50        0         0         0         0         0         50        17.30