2017-07-16 100 views
1

我想要獲得一個圖表來列出沿Y軸的A-Z,A在頂部,Z在底部(接近(0,0))。按鍵/值按字母順序排序列表

這是一個鵝卵石爛攤子,現在,因爲我正在學習如何做到這一點,但這裏的相關代碼:

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()) 

僅供參考這裏的print(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})

我想不通了解如何重新排列ie_count以按字母順序排列,使用鍵(字母)保留值(2417,171等)。

+0

我正在學習Python的,所以可能已經錯過了一些非常明顯的(如什麼名字/類型的我是什麼試圖做的是)。我會很感激任何關於這種情況的說明或降價原因。 – BruceWayne

+0

我沒有downvoted,而是'get_words'函數(有點不必要,因爲我們沒有訪問你的文件,所以你可以簡單地將它從問題中刪除),你可以插入你的'ie_count'而不是**和**函數'create_graph'會很有趣 – MSeifert

+1

@MSeifert - 啊,對不起!沒有意識到我離開了。我會編輯。僅供參考,來自GitHub的單詞列表是[此文件](https://github.com/dwyl/english-words/blob/master/words_alpha.txt)。 – BruceWayne

回答

2

您可以zip排序的.items並解壓鍵和值:

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) 
+0

哦,真是太棒了!我以前見過'zip()',但會更加註重它。最後一部分是如何向後重新排列。我嘗試了'reverse(letters)'/'reversed(vals)',但是得到「TypeError:'type'object'反向'沒有len()'。 – BruceWayne

+1

你的意思是」向後重新排列「是什麼意思?反序?這就像將'reverse = True'傳遞給'sorted'一樣簡單:'sorted(ie_count.items(),key = itemgetter(0),reverse = True)'或者只是使用「slicing」元組:'reversed_letters = letters [:: - 1]'和'reversed_vals = vals [:: - 1]' – MSeifert

+0

啊,那就做吧!非常感謝!:D(我加入了'reverse = True '到'sorted()'。 – BruceWayne