Question

我试图得到一个图表，在Y轴上列出A-Z，A在顶部，Z在底部（靠近(0,0)）。

这是一个乱七八糟的混乱，因为我正在学习如何做到这一点，但这是相关的代码：

from collections import Counter
import matplotlib.pyplot as plt; plt.rcdefaults()
import numpy as np
import matplotlib.pyplot as plt
from collections import OrderedDict
from operator import itemgetter

myFile = "D:/User/Documents/Data/English Words/words_alpha.txt"

ie_prefix = {}
ie_prefix.setdefault("letter", [])
ie_prefix.setdefault("word",[])
ie_prefix.setdefault("beforec",[])
ie_prefix_sorted = {}

def get_words(file_import):
    global total_words, ie_after_c, ie_after_not_c
  #if I don't do this and just keep the below, I get an error: 'total_words' is not defined when running the print(..., total_words, "total words and", ...) line.
    total_words = 0
    ie_after_not_c = 0
    ie_after_c = 0
    results = []
    with open(file_import) as inputfile:
        for line in inputfile:
            total_words = total_words + 1
            if line.find("ie") != -1:
                pos = line.find("ie")
                ie_prefix["letter"].append(line[pos-1:pos])
                ie_prefix["word"].append(line.strip('\n'))
                if line[pos-1:pos] == "c":
                    ie_prefix["beforec"].append(line.strip('\n'))
                    ie_after_c += 1
                elif line[pos-1:pos] != "c":
                    ie_after_not_c += 1

    ie_prefix_sorted = OrderedDict(sorted(ie_prefix.items()))
    return ie_prefix, total_words, ie_after_not_c, ie_after_c, ie_prefix_sorted


def create_graph(total_words, y_axis, x_axis):
    y_pos = np.arange(len(y_axis))
    fig, ax = plt.subplots()
    plt.barh(y_pos, x_axis, align='center', alpha=0.5)
    plt.yticks(y_pos, y_axis)
    plt.ylabel('Usage')
    plt.title('I before E rule')
    plt.legend()
    plt.show()

get_words(myFile)

# https://stackoverflow.com/questions/20316299/formatting-output-of-counter
ie_count = Counter(ie_prefix["letter"])

ie_count_sorted = sorted(ie_count)  #sorted(ie_count.items()) ## THis will just sort the KEYS I believe
ie_letters = list(ie_count_sorted)

###    
## How to use the SORTED IE Count in the graph, so it goes from A-Z where A is at the TOP, and Z is at the BOTTOM of Y-Axis (Z closest to (0,0))?  
create_graph(total, ie_count, ie_count.values())

仅供参考print(ie_count)：

反击（{'r'：2417，'t'：1771，'l'：1304，'f'：1034，'d'：778，'h'：765，'p'：753，'c '：729，'n'：647，'m'：536，'g'：492，'s'：470，'k'：443，'v'：273，'b'：260，'z'： 154，'你'：134，'w'：93，'o'：75，'x'：73，'y'：49，'e'：29，'a'：26，''：3，' j'：2，'我'：1}）

我无法弄清楚如何重新排列ie_count按字母顺序排列，保持值（2417,171等）与键（字母）。

Answer 1

您可以对.items进行排序，并使用zip解压键和值：

from collections import Counter
from operator import itemgetter

ie_count = Counter({'r': 2417, 't': 1771, 'l': 1304, 'f': 1034, 'd': 778, 'h': 765, 'p': 753, 'c': 729, 'n': 647, 'm': 536, 'g': 492, 's': 470, 'k': 443, 'v': 273, 'b': 260, 'z': 154, 'u': 134, 'w': 93, 'o': 75, 'x': 73, 'y': 49, 'e': 29, 'a': 26, '': 3, 'j': 2, 'i': 1})

cnts_sorted = sorted(ie_count.items(), key=itemgetter(0))
print(cnts_sorted)
# [('', 3), ('a', 26), ('b', 260), ('c', 729), ('d', 778), ('e', 29), 
#  ('f', 1034), ('g', 492), ('h', 765), ('i', 1), ('j', 2), ('k', 443), 
#  ('l', 1304), ('m', 536), ('n', 647), ('o', 75), ('p', 753), ('r', 2417), 
#  ('s', 470), ('t', 1771), ('u', 134), ('v', 273), ('w', 93), ('x', 73), 
#  ('y', 49), ('z', 154)]


letters, vals = zip(*cnts_sorted)
print(letters)
# ('', 'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n',
#  'o', 'p', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z')
print(vals)
# (3, 26, 260, 729, 778, 29, 1034, 492, 765, 1, 2, 443, 1304, 536, 647, 75,
# 753, 2417, 470, 1771, 134, 273, 93, 73, 49, 154)

按键/值按字母顺序对列表进行排序

1 个答案: